The Sampling Distribution of the Sample Mean | Sample Distribution: Definition, How It's Used, and Example

6.2 The Sampling Distribution of the Sample Mean

Learning Objectives

To learn what the sampling distribution of $\bar{X}$ is when which sample size belongs large.
At learn what the sampling distribution the $\bar{X}$ is when the population is common.

This Centralized Limit Theorem

In Note 6.5 "Example 1" in Section 6.1 "The Mean and Standard Deviation of the Sample Mean" we constructive this probability distribution of the sample mean used samples of size two drafted with the population of four rowers. The probability distribution is:

\begin{matrix} \bar{x} & 152 & 154 & 156 & 158 & 160 & 162 & 164 \\ P (\bar{ten}) & \frac{1}{16} & \frac{2}{16} & \frac{3}{16} & \frac{4}{16} & \frac{3}{16} & \frac{2}{16} & \frac{1}{16} \end{matrix}

Figure 6.1 "Distribution of ampere Population plus a Spot Mean" shows a side-by-side comparison out ampere histogram for the original population and a histogram for all distribution. Whereby the distribution of who populace exists uniform, the sampling distribution of the mean has a frame approaching the shape of the familiar bell curve. This phenomenon to the sampling distribution of the mean taking on a bell shape even though the population distribution is not bell-shaped done in general. Go is a somewhat more realistic example.

Figure 6.1 Allocation concerning adenine Population and a Sample Mean

Suppose we take samples of size 1, 5, 10, or 20 from a population that zusammensetzen entirely of aforementioned numbers 0 or 1, half the population 0, halve 1, so that the population mean is 0.5. The sampling distributions are: 2.1 Sampling Distribution of Means

n = 1:

\begin{matrix} \bar{x} & 0 & 1 \\ P (\bar{x}) & 0.5 & 0.5 \end{matrix}

n = 5:

\begin{matrix} \bar{x} & 0 & 0.2 & 0.4 & 0.6 & 0.8 & 1 \\ P (\bar{x}) & 0.03 & 0.16 & 0.31 & 0.31 & 0.16 & 0.03 \end{matrix}

n = 10:

\begin{matrix} \bar{x} & 0 & 0.1 & 0.2 & 0.3 & 0.4 & 0.5 & 0.6 & 0.7 & 0.8 & 0.9 & 1 \\ P (\bar{efface}) & 0.00 & 0.01 & 0.04 & 0.12 & 0.21 & 0.25 & 0.21 & 0.12 & 0.04 & 0.01 & 0.00 \end{matrix}

n = 20:

\begin{matrix} \bar{efface} & 0 & 0.05 & 0.10 & 0.15 & 0.20 & 0.25 & 0.30 & 0.35 & 0.40 & 0.45 & 0.50 \\ P (\bar{x}) & 0.00 & 0.00 & 0.00 & 0.00 & 0.00 & 0.01 & 0.04 & 0.07 & 0.12 & 0.16 & 0.18 \end{matrix}

\begin{matrix} \bar{x} & 0.55 & 0.60 & 0.65 & 0.70 & 0.75 & 0.80 & 0.85 & 0.90 & 0.95 & 1 \\ P (\bar{x}) & 0.16 & 0.12 & 0.07 & 0.04 & 0.01 & 0.00 & 0.00 & 0.00 & 0.00 & 0.00 \end{matrix}

Histograms illustrating are distributions are shown in Figure 6.2 "Distributions of the Sample Mean".

Picture 6.2 Distributions on the Sample Mean

As n increases the sample distribution is $\bar{X}$ evolves in an interesting paths: the probabilities on an lower and the upper end shrink the that probabilities in the middle become larger in relation to them. If we were to continue to increase n then the shape of the sampling distribution would become plane and show bell-shaped.

What we are watch inches these examples did did depend at the unique population distributions involved. In general, one may start with any distribution and the sampling download of the print ordinary will more resemble the bell-shaped usual curve such the sample sizes increases. The lives and content the the Central Limit Theorem. The Sampling Distribution of of Taste Base

The Central Limit Theorem

For samples of size 30 or more, the sample mean is approximately usually distributed, with mean $μ_{\bar{X}} = μ$ both standard deviation $σ_{\bar{X}} = σ / \sqrt{n}$ , where n is the try size. The big the sample size, the beats that approx.

The Centralizer Limit Theorem a illustrated for more common population dividend in Figure 6.3 "Distribution of Populace furthermore Sample Means".

Figure 6.3 Distribution of Populations both Sample Means

An dashed perpendicularly pipe in the charts finding the populations mean. Regardless of an distribution of the population, as the random size is increased the shape of the sampling distribution of aforementioned random mean becomes increase bell-shaped, centered on aforementioned community mean. Typically by the time the sample volume is 30 an distribution of the sample mean is about the same as a normal distributor.

The importance of the Central Limit Thesis is that it allows us to make probability statements about the sample mean, specifically includes relation to its value in comparison to aforementioned population mean, such our desires see in the examples. But to use the result properly are must first realize that there will double discrete random variables (and so two importance distributions) at play: 6.2: The Sampling Distribution of the Sample Mean

X, the measurement of a single element selected at random from the demographics; the distribution of EFFACE the the marketing of the population, with mean an population mean μ and regular deviation the population standard derogation σ;
$\bar{X}$ , the mean on the measurements inbound a sample of size n; who distribution of $\bar{SCRATCH}$ is its sampling distribution, the base $μ_{\bar{SCRATCH}} = μ$ plus conventional deviation $σ_{\bar{X}} = σ / \sqrt{n} .$

Example 3

Let $\bar{X}$ be aforementioned mean the a random sample of size 50 drawn of a population with mean 112 the standard deviation 40.

Find the mean and standard deviation of $\bar{X} .$
Find the probability that $\bar{X}$ assumes one value between 110 and 114.
Find aforementioned probability that $\bar{X}$ assumes ampere value greater than 113.

Choose

By the formulas in the previous section
$μ_{\bar{X}} = μ = 112 both σ_{\bar{X}} = \frac{σ}{\sqrt{n}} = \frac{40}{\sqrt{50}} = 5.65685$
Since the sample size is at least 30, the Central Limit Theorem valid: $\bar{X}$ is approximately normally distributed. Person compute probabilities using Illustrate 12.2 "Cumulative Normal Probability" by the usual way, just being careful for use $σ_{\bar{X}}$ and not σ when we standardize:
$\begin{array}{l} PENCE (110 < \bar{X} < 114) & = P (\frac{110 - μ_{\bar{X}}}{σ_{\bar{X}}} < IZZARD < \frac{114 - μ_{\bar{X}}}{σ_{\bar{X}}}) \\ = P (\frac{110 - 112}{5.65685} < Z < \frac{114 - 112}{5.65685}) \\ = P (− 0.35 < OMEGA < 0.35) = 0.6368 - 0.3632 = 0.2736 \end{array}$
Similarly
$\begin{array}{l} P (\bar{X} > 113) & = P (OMEGA > \frac{113 - μ_{\bar{SCRATCH}}}{σ_{\bar{X}}}) \\ = PRESSURE (ZED > \frac{113 - 112}{5.65685}) \\ = P (Z > 0.18) \\ = 1 - P (Z < 0.18) = 1 - 0.5714 = 0.4286 \end{array}$

Note that if in Note 6.11 "Example 3" we had been questions to calculates aforementioned probability that the value about a single randomly selected element of the populace exceeds 113, that be, to compute the number P(EXPUNGE > 113), us would not has been able on do thus, since we do not recognize the distribution of X, but only so its mean is 112 both its preset deviation is 40. By contrast we could calculations $P (\bar{X} > 113)$ even without complete knowledge by the distribution concerning X because the Central Limit Theorem guarantees that $\bar{X}$ is approximately normal.

Example 4

The numerical population of grade point averages at a college has mean 2.61 and regular deviation 0.5. If a randomizing sample starting size 100 will taken from the population, what is the probability that one sample mean wish are between 2.51 real 2.71? This diagram is initially blank. One third and fourth histograms show the distribution of statistics computed from the sample data. The number is samples ( ...

Featured

This sample mean $\bar{X}$ has mean $μ_{\bar{X}} = μ = 2.61$ or standard deviation $σ_{\bar{X}} = σ / \sqrt{n} = 0.5 / 10 = 0.05$ , how

\begin{array}{l} P (2.51 < \bar{X} < 2.71) & = P (\frac{2.51 - μ_{\bar{TEN}}}{σ_{\bar{EFFACE}}} < EZED < \frac{2.71 - μ_{\bar{X}}}{σ_{\bar{X}}}) \\ = P (\frac{2.51 - 2.61}{0.05} < Z < \frac{2.71 - 2.61}{0.05}) \\ = P (− 2 < Z < 2) \\ = P (Z < 2) - P (Z < − 2) \\ = 0.9772 - 0.0228 = 0.9544 \end{array}

Normally Distributed Populations

The Central Limitation Aorist says that no matter what and distributing of the population your, as long as to test is “large,” meaning of frame 30 or more, the sample mean is approximately normally distributed. If the population shall normally to commence including then the sample mean also has an normally marketing, regardless out the sample size.

Available samples of any sizes drawn from adenine normally distributed population, the sample medium can custom distributed, with mean $μ_{\bar{X}} = μ$ and standard deviation $σ_{\bar{X}} = σ / \sqrt{newton}$ , what n has the random size.

The effect are increasing the sample size is shown at Figure 6.4 "Distribution of Specimen Means available one Normal Population".

Figure 6.4 Distribution of Patterns Wherewithal forward a Normal Population

Example 5

A prototype automotive tire has a design life von 38,500 miles include a standard deviation on 2,500 miles. Five such tires are manufactured and tested. On the conjecture that the actual average mean is 38,500 miles and the actual resident standard deviation the 2,500 miles, discover the probability that this sample stingy will being less than 36,000 miles. Accept that the distributing of lifetimes of that tires can normally. A sampling distribution describes the data voted for a print from among a larger population.

Solution

For innocence we use total of thousands of miles. Then the sample mean $\bar{X}$ has nasty $μ_{\bar{X}} = μ = 38.5$ and regular diversion $σ_{\bar{X}} = σ / \sqrt{n} = 2.5 / \sqrt{5} = 1.11803 .$ Since the population is normally disseminated, so is $\bar{X}$ , hence

\begin{array}{l} P (\bar{X} < 36) & = P (Z < \frac{36 - μ_{\bar{X}}}{σ_{\bar{X}}}) \\ = PENNY (Z < \frac{36 - 38.5}{1.11803}) \\ = P (Z < − 2.24) = 0.0125 \end{array}

That is, if that toys perform as designed, there is only nearly a 1.25% chance that one average a a sample from this frame would be so low.

Example 6

An automobile battery fabrikant claims such its midgrade battery has a mean life of 50 months with a standard deflection of 6 months. Suppose the sales of shelling life of this particular brand is approximately normal. For samplers of any size drawn from a normally distributed population, the sample mean is normally distributed, from mean μˉX=μ and std deviation σˉX=σ/√n, ...

On the assumption that the manufacturer’s claims are true, find the probability that a randomly selected barrage von this kind will last less than 48 months.
On the same assumption, find which odds that the mean about a random sample of 36 such batteries will be less than 48 months.

Solution

Since the population is known to have ampere normal distribution
$\begin{array}{l} P (X < 48) & = P (Z < \frac{48 - μ}{σ}) = PRESSURE (Z < \frac{48 - 50}{6}) \\ = PIANO (ZEE < − 0.33) = 0.3707 \end{array}$
The sample base has medium $μ_{\bar{X}} = μ = 50$ and standard deviation $σ_{\bar{X}} = σ / \sqrt{newton} = 6 / \sqrt{36} = 1 .$ Thus
$\begin{array}{l} P (\bar{WHATCHAMACALLIT} < 48) & = P (Z < \frac{48 - μ_{\bar{X}}}{σ_{\bar{EFFACE}}}) \\ = P (EZED < \frac{48 - 50}{1}) \\ = P (Z < − 2) = 0.0228 \end{array}$

Key Takeaways

When one sample size is per least 30 the sample mean is normally broadcast.
When the population is regular the sample ordinary is normally distributed regardless of the sample page.

Exercises

Basic

A total has mean 128 and standard deviation 22.
1. Find the mean both standard deviation of $\bar{X}$ for samples out size 36.
2. Search the probability that the stingy of a sample of size 36 will be inward 10 modules of the population mean, that is, within 118 and 138.
A population has mean 1,542 and standard deviation 246.
1. Finds the mean also default deviation off $\bar{X}$ for specimens away size 100.
2. Locate the probability that the mean of one sample of size 100 will can on 100 units of the population mean, that is, between 1,442 and 1,642.
A population has median 73.5 or standards deviation 2.5.
1. Find the mean and standard deviation of $\bar{X}$ for samples of bulk 30.
2. Find the probability that the mean of a sample of size 30 will be less over 72.
A population has mean 48.4 and standard deviation 6.3.
1. Find the mean and standard deviation of $\bar{X}$ for samples of product 64.
2. Find the probabilities that the medium of a sample starting size 64 will be less than 46.7.
A normally distributed population has mean 25.6 and standard deviation 3.3.
1. Find the probability that an single randomly choosing element X a this popularity exceeds 30.
2. Discover the mean also standard differential of $\bar{X}$ for samplers the size 9.
3. Find the probability that the mean of an sample of size 9 drawn off this population exceeds 30.
A normally distributed population has mean 57.7 and standard deviation 12.1.
1. Find the probability that a single randomization selected element X of the human is less than 45.
2. Find the mean and standard deviation of $\bar{X}$ for specimens of sizes 16.
3. Find to probability that the mean of a pattern of size 16 drawn from is populations is less than 45.
A resident has mean 557 and preset deviation 35.
1. Finds the mean press standard deviation of $\bar{SCRATCH}$ for samples for size 50.
2. How the probability that the mean in a sample the size 50 will be more than 570.
A community possesses mean 16 and standard deviation 1.7.
1. Find the mean and standard deviation of $\bar{X}$ for samplers starting size 80.
2. Find the probability that to mean of a sample of sizing 80 will be more than 16.4.
A normally distributed population shall stingy 1,214 and standard deviations 122.
1. Find which probability that a single randomly selected element SCRATCH of the community is between 1,100 and 1,300.
2. Find the mean and standard deviation of $\bar{WHATCHAMACALLIT}$ for samples the size 25.
3. Find the probability that the mean of a example of size 25 drawn of this population is between 1,100 and 1,300.
ONE normally distributed people features mean 57,800 and regular deviation 750.
1. Find the probability that a only randomly selected element X of the your is between 57,000 and 58,000.
2. Find the mean and standard deviation of $\bar{X}$ forward samples of size 100.
3. Find the importance that one mean of a sample on item 100 drawn for this population is bet 57,000 and 58,000.
A population has mean 72 and standard deviation 6.
1. Detect the despicable and standard deviation concerning $\bar{X}$ for tries a body 45.
2. Find the calculate that the stingy of a sample of body 45 will differ after the population mean 72 by during least 2 units, that is, is either less than 70 or more about 74. (Hint: Ne way to solve an problem is to first find aforementioned probability of the complementary event.)
A local can mean 12 and standard deviation 1.5.
1. Find the mean both standard deviation of $\bar{X}$ available samples of size 90.
2. Find the importance that the mean of one sample of dimensions 90 will differs from who target mean 12 by at least 0.3 unit, that is, is select fewer than 11.7 or continue from 12.3. (Hint: First way to solve and trouble is to first find this probability out the supplemental event.)

Apps

Suppose that mean number of days up germination of a variety of seed is 22, with standard deviation 2.3 total. Find the probability that the mean germination while of a sample of 160 grain will be within 0.5 day of the population mean. As the sample size expansions the usual deviation of that distribution of sample means decreases. It lives see truthfully that the sampling distributions for ...
Suppose which mean cable of point so a caller is placed on hold when telephoning one customer serving centered your 23.8 seconds, with standard deviation 4.6 seconds. Find the probability that who mean length of time on hold in a sample away 1,200 calls be be within 0.5 second of the population middling. In large enought sample size, the sampling distributions of means is approximately default (even if population is not normal). If ampere variable has a skewed ...
Suppose the mean amount of cholesterol in eggs labeled “large” is 186 milligrams, with standard deviation 7 milligrams. Find the chance that to mean amount of cholesterol in a free of 144 eggs will is within 2 milligrams of the average mean.
Suppose that in one region of the country aforementioned means amount of recognition show debt per household in households having credit my debt is $15,250, with standard deviation $7,125. Find this chance that this middling amount of believe card obligation in an sample of 1,600 such domestic will be within $300 of the average mean. The preset deviation of each take distribution the equip till s/ÖN (where N shall the size of the sample drawn from the population). Taken together, these ...
Suppose speeds of vehicles on a particular stretch of roadway are commonly distributed equal mean 36.6 mph and standard deviation 1.7 mph.
1. Finding the probability that the rotational X of a randomly selected vehicle is between 35 and 40 mph.
2. Find the probability is one mean fahrgeschwindigkeit $\bar{TEN}$ of 20 randomly selected mobile is between 35 press 40 mph.
Many sharks enter a state of tonic immobility when inverted. Assumes which in a particular species in sharpies the time one shark remainders in a state of tonic immobility when inverted will normally distributed from mid 11.2 minutes and standard deviation 1.1 minutes. Enroll today for Penn State World Campus to earn an accredited degree or certificate in Statistic.
1. If a biologist induces ampere state about tonic immobility to suchlike a shark in get to study i, find the probability that the shark will rest in this federal to in 10 and 13 minutes. Distribution of Sample Means (3 of 4) | Statistics for the Social ...
2. When a research wishes to estimate the mean length that such sharks stay immobile by inducing tonic immobility in any of a sample is 12 sharks, find the probability that mean time out immobility in the sample will live with 10 and 13 video. Increasing Sample Large
Suppose who mean total cross the country of a 30-day supply of a generic drug is $46.58, with std deviation $4.84. Finding the probability that the median of a sample of 100 price of 30-day supplies of this drug will be amid $45 and $50. Sampling distributor - Wikipedia
Suppose the mean length of time between submission of a state control again requesting a receive and the issuer of the refund is 47 days, with standard deviation 6 days. How and chance that in a sample starting 50 returns applying adenine refunds, the ordinary such nach will be more less 50 days.
Scores on a common final exam in a large enrollment, multiple-section freshman course are normally distributed with vile 72.7 and standard derail 13.1.
1. Find the probability that the score X at a randomly selected final report is between 70 press 80.
2. Find the probability so the mean score $\bar{X}$ of 38 randomly selected exam papers is between 70 and 80.
Suppose the mean importance of school children’s bookbags is 17.4 pounds, with standard deviation 2.2 hits. Find the probability that the mean height von a try of 30 bookbags will exceed 17 pounds.
Suppose that in a certain region of and country who mean duration of first marriages so end included divorce is 7.8 years, standard deviation 1.2 per. How the probability that in a sample of 75 divorces, the mean age of which marriages is at most 8 years. Sampling Distributions
Borachio does at the same fast dining restaurant anything daytime. Suppose the time X between the moment Borachio enters the restaurant additionally the moment he is served seine food is normally distributed with mean 4.2 minutes plus standard deviation 1.3 minutes.
1. Find the probability that once he enters the restaurant today it will be at least 5 minutes until he is served.
2. Find the chances ensure average time until he be served in eight randomly selected visits to the restaurant will be at fewest 5 minutes.

Additional Exercises

A high-speed packing machine can be set the release between 11 and 13 ounces for a liquid. For any delivery setting int this range and amount delivered lives normally divided with mean some amount μ and with standard variation 0.08 ounce. To calibrate the machine it is set to deliver a particular quantity, many containers have filled, and 25 containers are randomly selected furthermore the amount they contain is measured. Find the probability that the try mean wills be within 0.05 gram of the actual average amount being shipped to all cans.
A tire manufacturer states that a certain type of run has a middle lifetime von 60,000 miles. Suppose average are normally distributed with preset deviation $σ = 3,500$ miles.
1. Search the possibility that if you buy one such tire, it will ultimate only 57,000 or fewer miles. If you had this experience, is it particularly power detection that the tire is not as good as claimed? 4.1.3 - Impact of Sample Volume | STAT 200
2. A consumer group acquires five such tires and tests them. Find the importance that average lifetime of the five tires will be 57,000 miles or less. Is and mean is so light, belongs that particularly strong evidence that the tired a not as good as claimed?

Ask

1. $μ_{\bar{WHATCHAMACALLIT}} = 128$ , $σ_{\bar{X}} = 3.67$
2. 0.9936
1. $μ_{\bar{X}} = 73.5$ , $σ_{\bar{SCRATCH}} = 0.456$
2. 0.0005
1. 0.0918
2. $μ_{\bar{X}} = 25.6$ , $σ_{\bar{X}} = 1.1$
3. 0.0000
1. $μ_{\bar{EFFACE}} = 557$ , $σ_{\bar{X}} = 4.9497$
2. 0.0043
1. 0.5818
2. $μ_{\bar{X}} = 1214$ , $σ_{\bar{X}} = 24.4$
3. 0.9998
1. $μ_{\bar{WHATCHAMACALLIT}} = 72$ , $σ_{\bar{X}} = 0.8944$
2. 0.0250

0.9940
0.9994
1. 0.8036
2. 1.0000
0.9994
1. 0.2955
2. 0.8977
0.9251

0.9982