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The central limit theorem

Why averages are normal: the sampling distribution of the mean.

Whatever the population looks like, the average of a large enough sample is approximately normal, centred on the population mean, with standard deviation σ/√n: the standard error. That single fact is what makes confidence intervals and hypothesis tests work.

The Central Limit Theorem for Sample Means (Averages)

Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution). Using a subscript that matches the random variable, suppose

${\mu }_{x}$ = the mean of X
${\sigma }_{x}$ = the standard deviation of X

If you draw random samples of size n, then as n increases, the random variable $\overline{X}$, which consists of sample means, tends to be normally distributed and

\[\overline{X}∼N({\mu }_{x},\frac{{\sigma }_{x}}{\sqrt{n}})\]

The central limit theorem for sample means says that if you keep drawing larger and larger samples (such as rolling one, two, five, and finally, ten dice) and calculating their means, the sample means form their own normal distribution (the sampling distribution). The normal distribution has the same mean as the original distribution and a variance that equals the original variance divided by the sample size. The variable n is the number of values that are averaged together, not the number of times the experiment is done.

To put it more formally, if you draw random samples of size n, the distribution of the random variable $\overline{X}$, which consists of sample means, is called the sampling distribution of the mean. The sampling distribution of the mean approaches a normal distribution as n, the sample size, increases.

The random variable $\overline{X}$ has a different z-score associated with it from that of the random variable X. The mean $\overline{x}$ is the value of $\overline{X}$ in one sample.

\[z=\frac{\overline{x}-{\mu }_{x}}{(\frac{{\sigma }_{x}}{\sqrt{n}})}\text{,}\]

μ_X is the average of both X and $\overline{X}$.

$\sigma \overline{x}=\frac{\sigma x}{\sqrt{n}}$ = standard deviation of $\overline{X}$ and is called the standard error of the mean.

Condensed: the full section is in OpenStax Statistics.

The Central Limit Theorem for Sample Means (Averages)

Use the following information to answer the next six exercises: Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours. Let Χ be the random variable representing the time it takes her to complete one Assume Χ is normally distributed. Let $\overline{X}$ be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews.

The Central Limit Theorem for Sums (Optional)

Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution) and suppose:

μ_X = the mean of Χ
σ_Χ = the standard deviation of X

If you draw random samples of size n, then as n increases, the random variable ΣX consisting of sums tends to be normally distributed and ΣΧ ~ N[(n)(μ_Χ), ($\sqrt{n}$)(σ_Χ)].

The central limit theorem for sums says that if you keep drawing larger and larger samples and taking their sums, the sums form their own normal distribution (the sampling distribution), which approaches a normal distribution as the sample size increases. The normal distribution has a mean equal to the original mean multiplied by the sample size and a standard deviation equal to the original standard deviation multiplied by the square root of the sample size.

The random variable ΣX has the following z-score associated with it:

Σx is one sum.
$z=\frac{\Sigma x-(n)({\mu }_{X})}{(\sqrt{n})({\sigma }_{X})}$
1. (n)(μ_X) = mean of ΣX
2. $(\sqrt{n})({\sigma }_{X})$ = standard deviation of $\Sigma X$

Example

An unknown distribution has a mean of 90 and a standard deviation of 15. A sample of size 80 is drawn randomly from the population.

Try it.

Find the probability that the sum of the 80 values (or the total of the 80 values) is more than 7,500.
Find the sum that is 1.5 standard deviations above the mean of the sums.

Solution

Let X = one value from the original unknown population. The probability question asks you to find a probability for the sum (or total of) 80 values.

ΣX = the sum or total of 80 values. Because μ_X = 90, σ_X = 15, and n = 80, $\Sigma X$ ~ N[(80)(90),
($\sqrt{\text{80}}$)(15)]

mean of the sums = (n)(μ_X) = (80)(90) = 7200
standard deviation of the sums = $\text{(}\sqrt{n}\text{)(}{\sigma }_{X}\text{) }=\text{ (}\sqrt{\text{80}}\text{)}$(15)
sum of 80 values = Σx = 7500

a. Find P(Σx > 7500)

P(Σx > 7500) = 0.0127

b. Find Σx where z = 1.5.

Σx = (n)(μ_X) + (z)$(\sqrt{n})$(σ_Χ) = (80)(90) + (1.5)($\sqrt{80}$)(15) = 7401.2

Condensed: the full section is in OpenStax Statistics.

The Central Limit Theorem for Sums (Optional)

Use the following information to answer the next four exercises: An unknown distribution has a mean of 80 and a standard deviation of 12. A sample size of 95 is drawn randomly from the population.

Use the following information to answer the next five exercises: The distribution of results from a cholesterol test has a mean of 180 and a standard deviation of 20. A sample size of 40 is drawn randomly.

Use the following information to answer the next six exercises: A researcher measures the amount of sugar in several cans of the same type of soda. The mean is 39.01 with a standard deviation of 0.5. The researcher randomly selects a sample of 100.

Use the following information to answer the next four exercises: An unknown distribution has a mean 12 and a standard deviation of one. A sample size of 25 is taken. Let X = the object of interest.

Use the following information to answer the next three exercises: A market researcher analyzes how many electronics devices customers buy in a single purchase. The distribution has a mean of three with a standard deviation of 0.7. She samples 400 customers.

Use the following information to answer the next three exercises: An unkwon distribution has a mean of 100, a standard deviation of 100, and a sample size of 100. Let X = one object of interest.

Using the Central Limit Theorem

It is important for you to understand when to use the central limit theorem. If you are being asked to find the probability of the mean, use the clt for the means. If you are being asked to find the probability of a sum or total, use the clt for sums. This also applies to percentiles for means and sums.

Examples of the Central Limit Theorem

The law of large numbers says that if you take samples of larger and larger sizes from any population, then the mean $\overline{x}$ of the samples tends to get closer and closer to μ. From the central limit theorem, we know that as n gets larger and larger, the sample means follow a normal distribution. The larger n gets, the smaller the standard deviation gets. (Remember that the standard deviation for $\overline{X}$ is $\frac{\sigma }{\sqrt{n}}$.) This means that the sample mean $\overline{x}$ must be close to the population mean μ. We can say that μ is the value that the sample means approach as n gets larger. The central limit theorem illustrates the law of large numbers.

Condensed: the full section is in OpenStax Statistics.

Using the Central Limit Theorem

Use the following information to answer the next 10 exercises: A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely, so the distribution of weights is uniform. A sample of 100 weights is taken.

Use the following information to answer the next five exercises: The length of time a particular smartphone's battery lasts follows an exponential distribution with a mean of ten months. A sample of 64 of these smartphones is taken.

Use the following information to answer the next six exercises: A uniform distribution has a minimum of six and a maximum of ten. A sample of 50 is taken.

Central limit theorem

In probability theory, the central limit theorem (CLT) states that, under appropriate conditions, the distribution of a normalized version of the sample mean converges to a standard normal distribution. This holds even if the original variables themselves are not normally distributed. There are several versions of the CLT, each applying in the context of different conditions.

The theorem is a key concept in probability theory because it implies that probabilistic and statistical methods that work for normal distributions can be applicable to many problems involving other types of distributions.

This theorem has seen many changes during the formal development of probability theory. Previous versions of the theorem date back to 1811, but in its modern form it was only precisely stated in the 1920s.

In statistics, the CLT can be stated as: let $X_1, X_2, \dots, X_n$ denote a statistical sample of size $n$ from a population with expected value (average) $\mu$ and finite positive variance $\sigma^2$, and let $\bar{X}_{n}$ denote the sample mean (which is itself a random variable). Then the limit as $n\to\infty$ of the distribution of $(\bar{X}_n-\mu) \sqrt{n}$ is a normal distribution with mean $0$ and variance $\sigma^2$.

In other words, suppose that a large sample of observations is obtained, each observation being randomly produced in a way that does not depend on the values of the other observations, and the average (arithmetic mean) of the observed values is computed. If this procedure is performed many times, resulting in a collection of observed averages, the central limit theorem says that if the sample size is large enough, the probability distribution of these averages will closely approximate a normal distribution.

The central limit theorem has several variants. In its common form, the random variables must be independent and identically distributed (i.i.d.). This requirement can be weakened; convergence of the mean to the normal distribution also occurs for non-identical distributions or for non-independent observations if they comply with certain conditions.

Condensed: the full section is in Wikipedia.

Classical CLT

Let $(X_n)_{n\geq 1}$ be a sequence of i.i.d. random variables having a distribution with expected value given by $\mu$ and finite variance given by $\sigma^2.$ Suppose we are interested in the sample average

\[\bar{X}_n \equiv \frac{X_1 + \cdots + X_n}{n}.\]

By the law of large numbers, the sample average converges almost surely (and therefore also converges in probability) to the expected value $\mu$ as $n\to\infty.$

The classical central limit theorem describes the size and the distributional form of the stochastic fluctuations around the deterministic number $\mu$ during this convergence. More precisely, it states that as $n$ gets larger, the distribution of the normalized mean $\sqrt{n}(\bar{X}_n - \mu)$, i.e. the difference between the sample average $\bar{X}_n$ and its limit $\mu,$ scaled by the factor $\sqrt{n}$, approaches the normal distribution with mean $0$ and variance $\sigma^2.$ For large enough $n,$ the distribution of $\bar{X}_n$ gets arbitrarily close to the normal distribution with mean $\mu$ and variance $\sigma^2/n.$

The usefulness of the theorem is that the distribution of $\sqrt{n}(\bar{X}_n - \mu)$ approaches normality regardless of the shape of the distribution of the individual $X_i.$ Formally, the theorem can be stated as follows:

Lindeberg-Lévy CLT, Suppose $X_1, X_2, X_3 \ldots$ is a sequence of i.i.d. random variables with $\operatorname E[X_i] = \mu$ and $\operatorname{Var}[X_i] = \sigma^2 < \infty.$ Then, as $n$ approaches infinity, the random variables $\sqrt{n}(\bar{X}_n - \mu)$ converge in distribution to a normal $\mathcal{N}(0, \sigma^2)$:

\[\sqrt{n}\left(\bar{X}_n - \mu\right) \mathrel{\overset{d}{\longrightarrow}} \mathcal{N}\left(0,\sigma^2\right) .\]

In the case $\sigma > 0,$ convergence in distribution means that the cumulative distribution functions of $\sqrt{n}(\bar{X}_n - \mu)$ converge pointwise to the cdf of the $\mathcal{N}(0, \sigma^2)$ distribution: for every real number $z,$

\[\lim_{n\to\infty} \mathbb{P}\left[\sqrt{n}(\bar{X}_n-\mu) \le z\right] = \lim_{n\to\infty} \mathbb{P}\left[\frac{\sqrt{n}(\bar{X}_n-\mu)}{\sigma } \le \frac{z}{\sigma}\right]= \Phi\left(\frac{z}{\sigma}\right) ,\]

Condensed: the full section is in Wikipedia.

Lyapunov CLT

In this variant of the central limit theorem the random variables $X_i$ have to be independent, but not necessarily identically distributed. The theorem also requires that random variables $\left| X_i\right|$ have moments of some order $(2+\delta)$, and that the rate of growth of these moments is limited by the Lyapunov condition given below.

Lyapunov CLT, Suppose $\{X_1, \ldots, X_n, \ldots\}$ is a sequence of independent random variables, each with finite expected value $\mu_i$ and variance $\sigma_i^2$. Define

\[s_n^2 = \sum_{i=1}^n \sigma_i^2 .\]

If for some $\delta > 0$, Lyapunov's condition

\[\lim_{n\to\infty} \; \frac{1}{s_{n}^{2+\delta}} \, \sum_{i=1}^{n} \operatorname E\left[\left|X_{i} - \mu_{i}\right|^{2+\delta}\right] = 0\]

is satisfied, then a sum of $\frac{X_i - \mu_i}{s_n}$ converges in distribution to a standard normal random variable, as $n$ goes to infinity:

\[\frac{1}{s_n}\,\sum_{i=1}^{n} \left(X_i - \mu_i\right) \mathrel{\overset{d}{\longrightarrow}} \mathcal{N}(0,1) .\]

In practice it is usually easiest to check Lyapunov's condition for $\delta = 1$.

If a sequence of random variables satisfies Lyapunov's condition, then it also satisfies Lindeberg's condition. The converse implication, however, does not hold.

Lindeberg (-Feller) CLT

In the same setting and with the same notation as above, the Lyapunov condition can be replaced with the following weaker one (from Lindeberg in 1920).

Suppose that for every $\varepsilon > 0$,

\[\lim_{n \to \infty} \frac{1}{s_n^2}\sum_{i = 1}^{n} \operatorname E\left[(X_i - \mu_i)^2 \cdot \mathbf{1}_{\left\{\left| X_i - \mu_i \right| > \varepsilon s_n \right\}} \right] = 0\]

where $\mathbf{1}_{\{\ldots\}}$ is the indicator function. Then the distribution of the standardized sums

\[\frac{1}{s_n}\sum_{i = 1}^n \left( X_i - \mu_i \right)\]

converges towards the standard normal distribution $\mathcal{N}(0, 1)$.

CLT for the sum of a random number of random variables

Rather than summing an integer number $n$ of random variables and taking $n \to \infty$, the sum can be of a random number $N$ of random variables, with conditions on $N$. For example, the following theorem is Corollary 4 of Robbins (1948). It assumes that $N$ is asymptotically normal (Robbins also developed other conditions that lead to the same result).

Robbins CLT, Let $\{X_i, i \geq 1\}$ be independent, identically distributed random variables with $E(X_i) = \mu$ and $\text{Var}(X_i) = \sigma^2$, and let $\{N_n, n \geq 1\}$ be a sequence of non-negative integer-valued random variables that are independent of $\{X_i, i \geq 1\}$. Assume for each $n = 1, 2, \dots$ that $E(N_n^2) < \infty$ and

\[\frac{N_n - E(N_n)}{\sqrt{\text{Var}(N_n)}} \xrightarrow{\quad d \quad} \mathcal{N}(0,1)\]

where $\xrightarrow{\,d\,}$ denotes convergence in distribution and $\mathcal{N}(0,1)$ is the normal distribution with mean 0, variance 1. Then

\[\frac{\sum_{i=1}^{N_n} X_i - \mu E(N_n)}{\sqrt{\sigma^2E(N_n) + \mu^2\text{Var}(N_n)}} \xrightarrow{\quad d \quad} \mathcal{N}(0,1)\]

Multidimensional CLT

Proofs that use characteristic functions can be extended to cases where each individual $\mathbf{X}_i$ is a random vector in $\R^k$, with mean vector $\boldsymbol\mu = \operatorname E[\mathbf{X}_i]$ and covariance matrix $\mathbf{\Sigma}$ (among the components of the vector), and these random vectors are independent and identically distributed. The multidimensional central limit theorem states that when scaled, sums converge to a multivariate normal distribution. Summation of these vectors is done component-wise.

For $i = 1, 2, 3, \ldots,$ let

\[\mathbf{X}_i = \begin{bmatrix} X_{i}^{(1)} \\ \vdots \\ X_{i}^{(k)} \end{bmatrix}\]

be independent random vectors. The sum of the random vectors $\mathbf{X}_1, \ldots, \mathbf{X}_n$ is

\[\sum_{i=1}^{n} \mathbf{X}_i = \begin{bmatrix} X_{1}^{(1)} \\ \vdots \\ X_{1}^{(k)} \end{bmatrix} + \begin{bmatrix} X_{2}^{(1)} \\ \vdots \\ X_{2}^{(k)} \end{bmatrix} + \cdots + \begin{bmatrix} X_{n}^{(1)} \\ \vdots \\ X_{n}^{(k)} \end{bmatrix} = \begin{bmatrix} \sum_{i=1}^{n} X_{i}^{(1)} \\ \vdots \\ \sum_{i=1}^{n} X_{i}^{(k)} \end{bmatrix}\]

and their average is

\[\mathbf{\bar X_n} = \begin{bmatrix} \bar X_{i}^{(1)} \\ \vdots \\ \bar X_{i}^{(k)} \end{bmatrix} = \frac{1}{n} \sum_{i=1}^{n} \mathbf{X}_i.\]

Theorem, Let $X_1, \dots, X_n, \dots$ be independent $\R^d$-valued random vectors, each having mean zero. Write $S =\sum^n_{i=1}X_i$ and assume $\Sigma = \operatorname{Cov}[S]$ is invertible. Let $Z \sim \mathcal{N}(0,\Sigma)$ be a $d$-dimensional Gaussian with the same mean and same covariance matrix as $S$. Then for all convex sets $U \subseteq \R^d$,

\[\left|\mathbb{P}[S \in U] - \mathbb{P}[Z \in U]\right| \le C \, d^{1/4} \gamma~,\] where $C$ is a universal constant, $\gamma = \sum^n_{i=1} \operatorname E \left[\left\| \Sigma^{-1/2}X_i\right\|^3_2\right]$, and $\|\cdot\|_2$ denotes the Euclidean norm on $\R^d$.

Condensed: the full section is in Wikipedia.

The generalized central limit theorem

The generalized central limit theorem (GCLT) was an effort of multiple mathematicians (Sergei Bernstein, Jarl Waldemar Lindeberg, Paul Lévy, William Feller, Andrey Kolmogorov, and others) over the period from 1920 to 1937. The first published complete proof of the GCLT was in 1937 by Paul Lévy in French. An English language version of the complete proof of the GCLT is available in the translation of Boris Vladimirovich Gnedenko and Kolmogorov's 1954 book.

The statement of the GCLT is as follows:

Statement of GCLT, A non-degenerate random variable Z is α-stable for some 0 < α ≤ 2 if and only if there is an independent, identically distributed sequence of random variables X₁, X₂, X₃, ..., and constants a_n > 0, b_n ∈ ℝ with \[a_n(X_1 + \dots + X_n) - b_n \to Z.\] Here, '→' means the sequence of random variable sums converges in distribution; i.e., the corresponding distributions satisfy F_n(y) → F(y) at all continuity points of F.

In other words, if sums of independent, identically distributed random variables converge in distribution to some Z, then Z must be a stable distribution.

CLT under weak dependence

A useful generalization of a sequence of independent, identically distributed random variables is a mixing random process in discrete time; "mixing" means, roughly, that random variables temporally far apart from one another are nearly independent. Several kinds of mixing are used in ergodic theory and probability theory. See especially strong mixing (also called α-mixing) defined by $\alpha(n) \to 0$ where $\alpha(n)$ is so-called strong mixing coefficient.

A simplified formulation of the central limit theorem under strong mixing is:

Theorem, Suppose that $\{X_1, \ldots, X_n, \ldots\}$ is stationary and $\alpha$-mixing with $\alpha_n = O\left(n^{-5}\right)$ and that $\operatorname E[X_n] = 0$ and $\operatorname E[X_n^{12}] < \infty$. Denote $S_n = X_1 + \cdots + X_n$, then the limit

\[\sigma^2 = \lim_{n\rightarrow\infty} \frac{\operatorname E\left(S_n^2\right)}{n}\]

exists, and if $\sigma \ne 0$ then $\frac{S_n}{\sigma\sqrt{n}}$ converges in distribution to $\mathcal{N}(0, 1)$.

In fact,

\[\sigma^2 = \operatorname E\left(X_1^2\right) + 2 \sum_{k=1}^{\infty} \operatorname E\left(X_1 X_{1+k}\right),\]

where the series converges absolutely.

The assumption $\sigma \ne 0$ cannot be omitted, since the asymptotic normality fails for $X_n = Y_n - Y_{n-1}$ where $Y_n$ are another stationary sequence.

There is a stronger version of the theorem: the assumption $\operatorname E\left[X_n^{12}\right] < \infty$ is replaced with $\operatorname E\left[{\left|X_n\right|}^{2+\delta}\right] < \infty$, and the assumption $\alpha_n = O\left(n^{-5}\right)$ is replaced with

Condensed: the full section is in Wikipedia.

Martingale difference CLT

Theorem, Let a martingale $M_n$ satisfy

$\frac1n \sum_{k=1}^n \operatorname E\left[\left(M_k-M_{k-1}\right)^2 \mid M_1,\dots,M_{k-1}\right] \to 1$ in probability as n → ∞,
for every ε > 0, $\frac1n \sum_{k=1}^n{\operatorname E\left[\left(M_k-M_{k-1} \right)^2\mathbf{1}\left[|M_k-M_{k-1}|>\varepsilon\sqrt{n}\right]\right]} \to 0$ as n → ∞,

then $\frac{M_n}{\sqrt{n}}$ converges in distribution to $\mathcal{N}(0, 1)$ as $n \to \infty$.

Proof of classical CLT

The central limit theorem has a proof using characteristic functions. It is similar to the proof of the (weak) law of large numbers.

Assume $\{X_1, \ldots, X_n, \ldots \}$ are independent and identically distributed random variables, each with mean $\mu$ and finite variance $\sigma^2$. The sum $X_1 + \cdots + X_n$ has mean $n\mu$ and variance $n\sigma^2$. Consider the random variable

\[Z_n = \frac{X_1+\cdots+X_n - n \mu}{\sqrt{n \sigma^2}} = \sum_{i=1}^n \frac{X_i - \mu}{\sqrt{n \sigma^2}} =: \sum_{i=1}^n \frac{1}{\sqrt{n}} Y_i,\]

where the last step defines the new random variables $Y_i := \frac{X_i - \mu}{\sigma}$, each with zero mean and unit variance ($\operatorname{var}(Y) = 1$). The characteristic function of $Z_n$ is given by

\[\begin{aligned} \varphi_{Z_n}\!(t) = \varphi_{\sum_{i=1}^n {\frac{1}{\sqrt{n}}Y_i}}\!(t) \ &=\ \varphi_{Y_1}\!\!\left(\frac{t}{\sqrt{n}}\right) \varphi_{Y_2}\!\! \left(\frac{t}{\sqrt{n}}\right)\cdots \varphi_{Y_n}\!\! \left(\frac{t}{\sqrt{n}}\right) \\[1ex] &=\ \left[\varphi_{Y_1}\!\!\left(\frac{t}{\sqrt{n}}\right)\right]^n, \end{aligned}\]

where the last step relies on the fact that all of the $Y_i$ are identically distributed. The characteristic function of $Y_1$ is, by Taylor's theorem, \[\varphi_{Y_1}\!\left(\frac{t}{\sqrt{n}}\right) = 1 - \frac{t^2}{2n} + o\!\left(\frac{t^2}{n}\right), \quad \left(\frac{t}{\sqrt{n}}\right) \to 0\]

where $o(t^2 / n)$ is "little o notation" for some function of $t$ that goes to zero more rapidly than $t^2 / n$. By the limit of the exponential function ($e^x = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n$), the characteristic function of $Z_n$ equals

Condensed: the full section is in Wikipedia.

Convergence to the limit

The central limit theorem gives only an asymptotic distribution. As an approximation for a finite number of observations, it provides a reasonable approximation only when close to the peak of the normal distribution; it requires a very large number of observations to stretch into the tails.

The convergence in the central limit theorem is uniform because the limiting cumulative distribution function is continuous. If the third central moment $\operatorname{E}\left[(X_1 - \mu)^3\right]$ exists and is finite, then the speed of convergence is at least on the order of $1 / \sqrt{n}$ (see Berry-Esseen theorem). Stein's method can be used not only to prove the central limit theorem, but also to provide bounds on the rates of convergence for selected metrics.

The convergence to the normal distribution is monotonic, in the sense that the entropy of $Z_n$ increases monotonically to that of the normal distribution.

The central limit theorem applies in particular to sums of independent and identically distributed discrete random variables. A sum of discrete random variables is still a discrete random variable, so that we are confronted with a sequence of discrete random variables whose cumulative probability distribution function converges towards a cumulative probability distribution function corresponding to a continuous variable (namely that of the normal distribution). This means that if we build a histogram of the realizations of the sum of n independent identical discrete variables, the piecewise-linear curve that joins the centers of the upper faces of the rectangles forming the histogram converges toward a Gaussian curve as n approaches infinity; this relation is known as de Moivre-Laplace theorem. The binomial distribution article details such an application of the central limit theorem in the simple case of a discrete variable taking only two possible values.

Common misconceptions

Studies have shown that the central limit theorem is subject to several common but serious misconceptions, some of which appear in widely used textbooks. These include:

The misconceived belief that the theorem applies to random sampling of any variable, rather than to the mean values (or sums) of iid random variables extracted from a population by repeated sampling. That is, the theorem assumes the random sampling produces a sampling distribution formed from different values of means (or sums) of such random variables.
The misconceived belief that the theorem ensures that random sampling leads to the emergence of a normal distribution for sufficiently large samples of any random variable, regardless of the population distribution. In reality, such sampling asymptotically reproduces the properties of the population, an intuitive result underpinned by the Glivenko-Cantelli theorem.
The misconceived belief that the theorem leads to a good approximation of a normal distribution for sample sizes greater than around 30, allowing reliable inferences regardless of the nature of the population. In reality, this empirical rule of thumb has no valid justification, and can lead to seriously flawed inferences. See Z-test for where the approximation holds.

Relation to the law of large numbers

The law of large numbers as well as the central limit theorem are partial solutions to a general problem: "What is the limiting behavior of S_n as n approaches infinity?" In mathematical analysis, asymptotic series are one of the most popular tools employed to approach such questions.

Suppose we have an asymptotic expansion of $f(n)$:

\[f(n)= a_1 \varphi_{1}(n)+a_2 \varphi_{2}(n)+O\big(\varphi_{3}(n)\big) \qquad (n \to \infty).\]

Dividing both parts by φ₁(n) and taking the limit will produce a₁, the coefficient of the highest-order term in the expansion, which represents the rate at which f(n) changes in its leading term.

\[\lim_{n\to\infty} \frac{f(n)}{\varphi_{1}(n)} = a_1.\]

Informally, one can say: "f(n) grows approximately as a₁φ₁(n)". Taking the difference between f(n) and its approximation and then dividing by the next term in the expansion, we arrive at a more refined statement about f(n):

\[\lim_{n\to\infty} \frac{f(n)-a_1 \varphi_{1}(n)}{\varphi_{2}(n)} = a_2 .\]

Condensed: the full section is in Wikipedia.

Calculating the variance

Let S_n be the sum of n random variables. Many central limit theorems provide conditions such that S_n/√Var(S_n) converges in distribution to N(0,1) (the normal distribution with mean 0, variance 1) as n → ∞. In some cases, it is possible to find a constant σ and function f(n) such that S_n/(σ√n⋅f(n)) converges in distribution to N(0,1) as n→ ∞.

Lemma, Suppose $X_1, X_2, \dots$ is a sequence of real-valued and strictly stationary random variables with $\operatorname E(X_i) = 0$ for all $i$, $g : [0,1] \to \R$, and $S_n = \sum_{i=1}^{n} g\left(\tfrac{i}{n}\right) X_i$. Construct

\[\sigma^2 = \operatorname E(X_1^2) + 2\sum_{i=1}^\infty \operatorname E(X_1 X_{1+i})\]

If $\sum_{i=1}^\infty \operatorname E(X_1 X_{1+i})$ is absolutely convergent, $\left| \int_0^1 g(x)g'(x) \, dx\right| < \infty$, and $0 < \int_0^1 (g(x))^2 dx < \infty$ then $\mathrm{Var}(S_n)/(n \gamma_n) \to \sigma^2$ as $n \to \infty$ where $\gamma_n = \frac{1}{n}\sum_{i=1}^{n} \left(g\left(\tfrac{i}{n}\right)\right)^2$.
If in addition $\sigma > 0$ and $S_n/\sqrt{\mathrm{Var}(S_n)}$ converges in distribution to $\mathcal{N}(0,1)$ as $n \to \infty$ then $S_n/(\sigma\sqrt{n \gamma_n})$ also converges in distribution to $\mathcal{N}(0,1)$ as $n \to \infty$.

Вработен пример · 10/sqrt(25)

Arithmetic · Arithmetic

Evaluate 10/sqrt(25)

\frac{10}{\sqrt{25}}

Чекор по чекор

\frac{10}{\sqrt{25}} = 10 \cdot \frac{1}{5}
Power: 1/(√(25)) = 1/5.
10 \cdot \frac{1}{5} = 2
Multiply: 10(1/5) = 2.

Одкриј го одговорот

Сега ти. Секој чекор, слика, одговорот е скриен додека не прашаш.

Вежбај (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

a. Find the probability that the sample mean is between 85 and 92.

Одкриј го одговорот

a. Let X = one value from the original unknown population. The probability question asks you to find a probability for the sample mean.
Let $\overline{X}$ = the mean of a sample of size 25. Because μ_x = 90, σ_x = 15, and n = 25,
\[\overline{X}∼N({\mu }_{x},\frac{{\sigma }_{x}}{\sqrt{n}})\]
Find P(85 < $\overline{x}$ < 92). Draw a graph.
P(85 < $\overline{x}$ < 92) = 0.6997
The probability that the sample mean is between 85 and 92 is 0.6997.
b. Find the value that is two standard deviations above the expected value, 90, of the sample mean.

Одкриј го одговорот

b. To find the value that is two standard deviations above the expected value 90, use the following formula
\[\text{value}={µ}_{x}+(\#\text{ofSTDEVs})(\frac{{\sigma }_{x}}{\sqrt{n}})\]\[\text{value }=\text{ 90 }+\text{ 2 }(\frac{15}{\sqrt{25}})=96\text{.}\]
The value that is two standard deviations above the expected value is 96.
The standard error of the mean is $\frac{\sigma x}{\sqrt{n}}$ = $\frac{15}{\sqrt{25}}$ = 3. Recall that the standard error of the mean is a description of how far (on average) that the sample mean will be from the population mean in repeated simple random samples of size n.
An unknown distribution has a mean of 45 and a standard deviation of eight. Samples of size n = 30 are drawn randomly from the population. Find the probability that the sample mean is between 42 and 50.

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P(42 < $\overline{x}$ < 50) = $(\text{42,50,45,}\frac{8}{\sqrt{30}})$ = 0.9797
The length of time, in hours, it takes a group of people, 40 years old and older, to play one soccer match is normally distributed with a mean of 2 hours and a standard deviation of 0.5 hours. A sample of size n = 50 is drawn randomly from the population. Find the probability that the sample mean is between 1.8 hours and 2.3 hours.

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Let X = the time, in hours, it takes to play one soccer match.
The probability question asks you to find a probability for the sample mean time, in hours, it takes to play one soccer match.
Let $\overline{X}$ = the mean time, in hours, it takes to play one soccer match.
If μ_X = _________, σ_X = __________, and n = ___________, then X ~ N(______, ______) by the central limit theorem for means.
μ_X = 2, σ_X = 0.5, n = 50, and X ~ N$(\text{2, }\frac{0.5}{\sqrt{50}})$
Find P(1.8 < $\overline{x}$ < 2.3). Draw a graph.
\[P(1.8<\overline{x}<2.3)=0.9977\]
normalcdf\[(1.\text{8,2}\text{.3,2,}\frac{.5}{\sqrt{50}})=\text{ 0.9977}\]
The probability that the mean time is between 1.8 hours and 2.3 hours is 0.9977.
The length of time taken on the SAT exam for a group of students is normally distributed with a mean of 2.5 hours and a standard deviation of 0.25 hours. A sample size of n = 60 is drawn randomly from the population. Find the probability that the sample mean is between two hours and three hours.

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P(2 < $\overline{x}$ < 3) = normalcdf$(\text{2, 3, 2}\text{.5, }\frac{0.25}{\sqrt{60}})$ = 1
In a recent study reported Oct. 29, 2012, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. Take a sample of size n = 100.
1. What are the mean and standard deviation for the sample mean ages of tablet users?
2. What does the distribution look like?
3. Find the probability that the sample mean age is more than 30 years (the reported mean age of tablet users in this particular study).
4. Find the 95^th percentile for the sample mean age (to one decimal place).
Одкриј го одговорот
1. Because the sample mean tends to target the population mean, we have μ_χ = μ = 34. The sample standard deviation is given by ${\sigma }_{\text{\chi }}$ = $\frac{\sigma }{\sqrt{n}}$ = $\frac{15}{\sqrt{100}}$ = $\frac{15}{10}$ = 1.5.
2. The central limit theorem states that for large sample sizes (n), the sampling distribution will be approximately normal.
3. The probability that the sample mean age is more than 30 is given by P(Χ > 30) = normalcdf(30,E99,34,1.5) = 0.9962.
4. Let k = the 95^th percentile.
  k = invNorm$(0.\text{95,34,}\frac{15}{\sqrt{100}})$ = 36.5
A gaming marketing gap for men between the ages of 30 to 40 has been identified. You are researching a startup game targeted at the 35-year-old demographic. Your idea is to develop a strategy game that can be played by men from their late 20s through their late 30s. Based on the article’s data, industry research shows that the average strategy player is 28 years old with a standard deviation of 4.8 years. You take a sample of 100 randomly selected gamers. If your target market is 29- to 35-year-olds, should you continue with your development strategy?

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You need to determine the probability for men whose mean age is between 29 and 35 years of age wanting to play a strategy game.
P(29 < $\overline{x}$ < 35) = normalcdf$(\text{29,35,28,}\frac{4.8}{\sqrt{100}})$ = 0.0186
You can conclude there is approximately a 1.9% chance that your game will be played by men whose mean age is between 29 and 35.
The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of 60.
1. What are the mean and standard deviation for the sample mean number of app engagement minutes by a tablet user?
2. What is the standard error of the mean?
3. Find the 90^th percentile for the sample mean time for app engagement for a tablet user. Interpret this value in a complete sentence.
4. Find the probability that the sample mean is between eight minutes and 8.5 minutes.
Одкриј го одговорот
1. ${\mu }_{\overline{x}}=\mu =8.2\ {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}=\frac{1}{\sqrt{60}}=0.13$
2. This allows us to calculate the probability of sample means of a particular distance from the mean, in repeated samples of size 60.
3. Let k = the 90^th percentile.
  k = invNorm$(0.\text{90,8}\text{.2,}\frac{1}{\sqrt{60}})$ = 8.37. This values indicates that 90 percent of the average app engagement time for table users is less than 8.37 minutes.
4. P(8 < $\overline{x}$ < 8.5) = normalcdf$(\text{8,8}\text{.5,8}\text{.2,}\frac{1}{\sqrt{60}})$ = 0.9293
Cans of a cola beverage claim to contain 16 ounces. The amounts in a sample are measured and the statistics are n = 34, $\overline{x}$ = 16.01 ounces. If the cans are filled so that μ = 16.00 ounces (as labeled) and σ = 0.143 ounces, find the probability that a sample of 34 cans will have an average amount greater than 16.01 ounces. Do the results suggest that cans are filled with an amount greater than 16 ounces?

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We have P($\overline{x}$ > 16.01) = normalcdf$(\text{16}\text{.01,E99,16,}\frac{0.143}{\sqrt{34}})$ = 0.3417. Since there is a 34.17% probability that the average sample weight is greater than 16.01 ounces, we should be skeptical of the company’s claimed volume. If I am a consumer, I should be glad that I am probably receiving free cola. If I am the manufacturer, I need to determine if my bottling processes are outside of acceptable limits.
What is the mean, standard deviation, and sample size?

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mean = 4 hours, standard deviation = 1.2 hours, sample size = 16
Complete the distributions.
1. X ~ _____(_____, _____)
2. $\overline{X}$ ~ _____(_____, _____)
Find the probability that one review will take Yoonie from 3.5 to 4.25 hours. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.
2. P(________ < x < ________) = _______
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a. Check student's solution.
b. 3.5, 4.25, 0.2441
Find the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hrs. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.
2. P(________________) = _______
What causes the probabilities in and to be different?

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The fact that the two distributions are different accounts for the different probabilities.
Find the 95^th percentile for the mean time to complete one month's reviews. Sketch the graph.
2. The 95^th percentile =____________
Previously, De Anza's statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $0.88. Suppose that we randomly pick 25 daytime statistics students.
1. In words, Χ = ____________.
2. Χ ~ _____(_____, _____)
3. In words, $\overline{X}$ = ____________.
4. $\overline{X}$ ~ ______ (______, ______)
5. Find the probability that an individual had between $0.80 and $1.00. Graph the situation, and shade in the area to be determined.
6. Find the probability that the average amount of change of the 25 students was between $0.80 and $1.00. Graph the situation, and shade in the area to be determined.
7. Explain why there is a difference in part (e) and part (f).
Одкриј го одговорот
1. Χ = amount of change students carry
2. Χ ~ Exp(1/0.88) or approximately Χ ~ Exp(1.1364)
3. $\overline{X}$ = average amount of change carried by a sample of 25 students.
4. $\overline{X}$ ~ N(0.88, 0.176)
5. 0.0819
6. 0.4276
7. The probability in part (e) represents the probability of an individual value. In part (f), the probability describes the mean of a sample of 25. Part (f) relies on the central limit theorem, so the distributions are different. Part (e) is exponential and part (f) is normal.
Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls.
1. If $\overline{X}$ = average distance in feet for 49 fly balls, then $\overline{X}$ ~ _______(_______, _______).
2. What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the horizontal axis for $\overline{X}$. Shade the region corresponding to the probability. Find the probability.
3. Find the 80^th percentile of the distribution of the average of 49 fly balls.
According to the Internal Revenue Service, the average length of time for an individual to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040 is 10.53 hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is two hours. Suppose we randomly sample 36 taxpayers.
1. In words, Χ = _____________.
2. In words, $\overline{X}$ = _____________.
3. $\overline{X}$ ~ _____(_____, _____)
4. Would you be surprised if the 36 taxpayers finished their Form 1040s in an average of more than 12 hours? Explain why or why not in complete sentences.
5. Would you be surprised if one taxpayer finished his or her Form 1040 in more than 12 hours? In a complete sentence, explain why.
Одкриј го одговорот
1. length of time for an individual to complete IRS form 1040, in hours
2. mean length of time for a sample of 36 taxpayers to complete IRS form 1040, in hours
3. N$(\text{10}\text{.53, }\frac{1}{3})$
4. Yes, I would be surprised, because the probability is almost 0.
5. No, I would not be totally surprised because the probability is 0.2312.
Suppose that a category of world-class runners are known to run a marathon (26 miles) in an average of 145 minutes with a standard deviation of 14 minutes. Consider 49 of the races. Let $\overline{X}$ be the average of the 49 races.
1. $\overline{X}$ ~ _____(_____, _____)
2. Find the probability that the runner will average between 142 and 146 minutes in these 49 marathons.
3. Find the 80^th percentile for the average of these 49 marathons.
4. Find the median of the average running times.
The length of songs in a collector’s online album collection is uniformly distributed from 2 to 3.5 minutes. Suppose we randomly pick five albums from the collection. There are a total of 43 songs on the five albums.
1. In words, Χ = _________.
2. Χ ~ _____________
3. In words, $\overline{X}$ = _____________.
4. $\overline{X}$ ~ _____(_____, _____)
5. Find the first quartile for the average song length.
6. The IQR for the average song length is _______, _______.
Одкриј го одговорот
1. the length of a song, in minutes, in the collection
2. U(2, 3.5)
3. the average length, in minutes, of the songs from a sample of five albums from the collection
4. N(2.75, 0.0220)
5. 2.74 minutes
6. 0.03 minutes
In 1940, the average size of a U.S. farm was 174 acres. Let’s say that the standard deviation was 55 acres. Suppose we randomly survey 38 farmers from 1940.
1. In words, Χ = _____________.
2. In words, $\overline{X}$ = _____________.
3. $\overline{X}$ ~ _____(_____, _____)
4. The IQR for $\overline{X}$ is from _______ acres to _______ acres.
Determine which of the following are true and which are false. Then, in complete sentences, justify your answers.
1. When the sample size is large, the mean of $\overline{X}$ is approximately equal to the mean of Χ.
2. When the sample size is large, $\overline{X}$ is approximately normally distributed.
3. When the sample size is large, the standard deviation of $\overline{X}$ is approximately the same as the standard deviation of Χ.
Одкриј го одговорот
1. True. The mean of a sampling distribution of the means is approximately the mean of the data distribution.
2. True. According to the central limit theorem, the larger the sample, the closer the sampling distribution of the means becomes normal.
3. The standard deviation of the sampling distribution of the means will decrease, making it approximately the same as the standard deviation of X as the sample size increases.
The percentage of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of about ten. Suppose that 16 individuals are randomly chosen. Let $\overline{X}$ = average percentage of fat calories.
1. $\overline{X}$ ~ ______(______, ______)
2. For the group of 16, find the probability that the average percentage of fat calories consumed is more than five. Graph the situation and shade in the area to be determined.
3. Find the first quartile for the average percentage of fat calories.
The distribution of income in some economically developing countries is considered wedge shaped (many very poor people, very few middle income people, and even fewer wealthy people). Suppose we pick a country with a wedge-shaped distribution. Let the average salary be $2,000 per year with a standard deviation of $8,000. We randomly survey 1,000 residents of that country.
1. In words, Χ = _____________.
2. In words, $\overline{X}$ = _____________.
3. $\overline{X}$ ~ _____(_____, _____)
4. How is it possible for the standard deviation to be greater than the average?
5. Why is it more likely that the average salary of the 1,000 residents will be from $2,000 to $2,100 than from $2,100 to $2,200?
Одкриј го одговорот
1. X = the yearly income of someone in a Third World country
2. the average salary from samples of 1,000 residents of a Third World country
3. $\overline{X}$ ∼ N$(\text{2,000, }\frac{\text{8,000}}{\sqrt{\text{1,000}}})$
4. Very wide differences in data values can have averages smaller than standard deviations.
5. The distribution of the sample mean will have higher probabilities closer to the population mean.
  P(2,000 < $\overline{X}$ < 2,100) = 0.1537
  P(2,100 < $\overline{X}$ < 2,200) = 0.1317
Which of the following is NOT true about the distribution for averages?
1. The mean, median, and mode are equal.
2. The area under the curve is 1.
3. The curve never touches the x-axis.
4. The curve is skewed to the right.
The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations. The distribution to use for the average cost of gasoline for the 16 gas stations is:
1. $\overline{X}$ ~ N(4.59, 0.10)
2. $\overline{X}$ ~ N$(\text{4}\text{.59, }\frac{0.10}{\sqrt{16}})$
3. $\overline{X}$ ~ N$(\text{4}\text{.59, }\frac{16}{0.10})$
4. $\overline{X}$ ~ N$(\text{4}\text{.59, }\frac{\sqrt{16}}{0.10})$
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b
1. Find the probability that the sum of the 80 values (or the total of the 80 values) is more than 7,500.
2. Find the sum that is 1.5 standard deviations above the mean of the sums.
Одкриј го одговорот
Let X = one value from the original unknown population. The probability question asks you to find a probability for the sum (or total of) 80 values.
ΣX = the sum or total of 80 values. Because μ_X = 90, σ_X = 15, and n = 80, $\Sigma X$ ~ N[(80)(90),
($\sqrt{\text{80}}$)(15)]
- mean of the sums = (n)(μ_X) = (80)(90) = 7200
- standard deviation of the sums = $\text{(}\sqrt{n}\text{)(}{\sigma }_{X}\text{) }=\text{ (}\sqrt{\text{80}}\text{)}$(15)
- sum of 80 values = Σx = 7500
a. Find P(Σx > 7500)
P(Σx > 7500) = 0.0127
Note
normalcdf(lower value, upper value, mean of sums, stdev of sums)
The parameter list is abbreviated(lower, upper, (n)(μ_X, $(\sqrt{n})$(σ_X))
normalcdf (7500,1E99,(80)(90),$(\sqrt{80})$(15)) = 0.0127
Reminder
1E99 = 10⁹⁹.
Press the EE key for E.
b. Find Σx where z = 1.5.
Σx = (n)(μ_X) + (z)$(\sqrt{n})$(σ_Χ) = (80)(90) + (1.5)($\sqrt{80}$)(15) = 7401.2
An unknown distribution has a mean of 45 and a standard deviation of 8. A sample size of 50 is drawn randomly from the population. Find the probability that the sum of the 50 values is more than 2,400.

Одкриј го одговорот

0.0040
In a recent study reported Oct. 29, 2012, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. The sample size is 50.
1. What are the mean and standard deviation for the sum of the ages of tablet users? What is the distribution?
2. Find the probability that the sum of the ages is between 1,500 and 1,800 years.
3. Find the 80^th percentile for the sum of the 50 ages.
Одкриј го одговорот
1. μ_Σx = nμ_x = 50(34) = 1,700 and σ_Σx = $\sqrt{n}$σ_x = $(\sqrt{\text{50}}\text{ )}$(15) = 106.01
  The distribution is normal for sums by the central limit theorem.
2. P(1500 < Σx < 1800) = normalcdf (1500, 1800, (50)(34), $(\sqrt{\text{50}}\text{ )}$(15)) = 0.7974
3. Let k = the 80^th percentile.
  k = invNorm(0.80,(50)(34),$(\sqrt{\text{50}}\text{ )}$(15)) = 1789.3
In a recent study reported Oct.29, 2012, the mean age of tablet users is 35 years. Suppose the standard deviation is 10 years. The sample size is 39.
1. What are the mean and standard deviation for the sum of the ages of tablet users? What is the distribution?
2. Find the probability that the sum of the ages is between 1,400 and 1,500 years.
3. Find the 90^th percentile for the sum of the 39 ages.
Одкриј го одговорот
1. μ_Σx = nμ_x = 1,365 and σ_Σx = $\sqrt{n}{\sigma }_{x}$ = 62.4
  The distribution is normal for sums by the central limit theorem.
2. P(1400 < Σ_x < 1500) = normalcdf (1400,1500,(39)(35),($\sqrt{39}$)(10)) = 0.2723
3. Let k = the 90^th percentile.
  k = invNorm(0.90,(39)(35),($\sqrt{39}$) (10)) = 1445.0
The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample size of 70.
1. What are the mean and standard deviation for the sums?
2. Find the 95^th percentile for the sum of the sample. Interpret this value in a complete sentence.
3. Find the probability that the sum of the sample is at least 10 hours.
Одкриј го одговорот
1. μ_Σx = nμ_x = 70(8.2) = 574 minutes and σ_Σx = $(\sqrt{n})({\sigma }_{x})$ = $(\sqrt{\text{70}}\text{ )}$(1) = 8.37 minutes
2. Let k = the 95^th percentile.
  k = invNorm (0.95,(70)(8.2),$(\sqrt{\text{70}}\text{)}$(1)) = 587.76 minutes
  Ninety-five percent of the app engagement times are at most 587.76 minutes.
3. 10 hours = 600 minutes
  P(Σx ≥ 600) = normalcdf(600,E99,(70)(8.2),$(\sqrt{\text{70}}\text{)}$(1)) = 0.0009
The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample size of 70.
1. What is the probability that the sum of the sample is between seven hours and 10 hours? What does this mean in context of the problem?
2. Find the 84^th and 16^th percentiles for the sum of the sample. Interpret these values in context.
Одкриј го одговорот
1. 7 hours = 420 minutes
  10 hours = 600 minutes
  normalcdf$P(420\le \Sigma x\le 600)=normalcdf(420,600,(70)(8.2),\sqrt{70}(1))=0.9991$
  This means that for this sample sums there is a 99.9% chance that the sums of usage minutes will be between 420 minutes and 600 minutes.
2. $invNorm(0.84,(70)(8.2),\sqrt{70}(1))=582.32$
  $invNorm(0.16,(70)(8.2),\sqrt{70}(1))=565.68$
  Since 84% of the app engagement times are at most 582.32 minutes and 16% of the app engagement times are at most 565.68 minutes, we may state that 68% of the app engagement times are between 565.68 minutes and 582.32 minutes.
Find the probability that the sum of the 95 values is greater than 7,650.

Одкриј го одговорот

0.3345
Find the probability that the sum of the 95 values is less than 7,400.
Find the sum that is two standard deviations above the mean of the sums.

Одкриј го одговорот

7833.92
Find the sum that is 1.5 standard deviations below the mean of the sums.
Find the probability that the sum of the 40 values is greater than 7,500.

Одкриј го одговорот

0.0089
Find the probability that the sum of the 40 values is less than 7,000.
Find the sum that is one standard deviation above the mean of the sums.

Одкриј го одговорот

7326.49
Find the sum that is 1.5 standard deviations below the mean of the sums.

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Потпиши се Најава

Симболи кои се користат тука

Допрете било кој симбол за целосната дефиниција, слика, и што значи секоја буква во неа.

Како да го направиш тоа?: The central limit theorem

Power: 1/(√(25)) = 1/5.
Multiply: 10(1/5) = 2.

Прашања што луѓето ги поставуваат

Mean or median, which should I use?

Median when the data have outliers or a long tail (incomes, house prices); mean when the data are roughly symmetric and you want every value to count. Report both if they disagree. The gap is itself information.

What does a p-value actually say?

The probability of seeing data at least this extreme if the null hypothesis were true. It is not the probability that the null hypothesis is true.

Why divide by n − 1 for the sample variance?

The sample mean sits closer to the sample than the true mean does, so squared deviations from it are slightly too small on average; dividing by n − 1 instead of n corrects the bias.

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