1. 7.3 Sample Means
HW: p. 454 (49-63 odd, 65-68)

2. Mean & Standard Deviation of the Sampling Distribution of 𝑥
Suppose that 𝑥 is the mean of a sample from a large population with mean 𝜇 and standard deviation 𝜎. Then the mean and standard deviation of the sampling distribution are:
Mean: 𝜇 𝑥 =𝜇 Standard Deviation: 𝜎 𝑥 = 𝜎 𝑛
*larger samples mean less variability *as long as the 10% condition is satisfied

3. Example: Movie-going Students
Suppose that the number of movies viewed in the last year by high school students has an average of 19.3 with a standard deviation of Suppose we take an SRS of 100 high school students and calculate the mean number of movies viewed by the members of the sample.
What is the mean of the sampling distribution of 𝑥 ?
𝜇 𝑥 =𝜇=19.3
What is the standard deviation of the sampling distribution of 𝑥 ? Check whether the 10% condition is satisfied.
𝜎 𝑥 = 𝜎 𝑛 = =1.58
The 10% condition is met because there are at least 10(100) = 1000 high school
students.
𝜇 𝑥 =𝜇=19.3
𝜎 𝑥 = 𝜎 𝑛 = =1.58% The 10% condition is met because there are at least 10(100) = 1000 high school students.

4. Example: 5th Grade Math
The times it takes 5th graders to complete a particular mathematics problem are Normally distributed with mean 2 minutes and standard deviation 0.8 minutes.
Find the probability that a randomly chosen 5th grader will take more than minutes to complete the problem. Show your work.
𝜇=2 Sketch:
𝜎=0.8
We want to find 𝑃(𝑥>2.5)
𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 2.5, 1× , 2, 0.8 =0.266
𝑧= 2.5−2 .8 = 𝑃 𝑥>0.625 =0.2659
𝑧= 2.5− = 𝑃 𝑥 >2.5 =

5. Example: 5th Grade Math
The times it takes 5th graders to complete a particular mathematics problem are Normally distributed with mean 2 minutes and standard deviation 0.8 minutes.
Suppose you give the problem to an SRS of 20 students. Sketch the sampling distribution of 𝑥 . Then use the distribution to determine the probability that the mean time to complete the problem for the SRS of students is greater than 2.5 minutes. Show your work.
𝜇 𝑥 = Sketch:
𝜎 𝑥 = =0.179
We want to find 𝑃( 𝑥 >2.5)
𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 2.5, 1× , 2, =0.0026
𝑧= 2.5−2 .8 = 𝑃 𝑥>0.625 =0.2659
𝑧= 2.5− = 𝑃 𝑥 >2.5 =

6. Shape:
The shape of the sampling distribution will depend upon the shape of the population distribution.
If the population is Normally distributed, the sampling distribution of 𝑥 will be Normally distributed.
If the population distribution is non-Normal, the sampling distribution of 𝑥 will become more and more Normal as n increases.

7. Central Limit Theorem
Most population distributions are not Normally distributed, which means we could not use Normal calculations.
Thanks to the Central Limit Theorem, when the sample size n is large, the shape of the sampling distribution of 𝑥 will be approximately Normal no matter what shape the population distribution may be!
We define “large” to be any sample that is at least 30.
So, if 𝑛≥30, we can be safe in assuming that the sampling distribution is Normal and use Normal calculations.
If 𝑛<30, we must consider the shape of the population distribution.

8. Example: Cholesterol
The blood cholesterol level of adult men has mean 188 mg/dl and standard deviation 41 mg/dl. A SRS of 250 men is selected and the mean blood cholesterol level in the sample is calculated.
Sketch the sampling distribution of 𝑥 and calculate the probability that the sample mean will be greater than 193.
𝜇 𝑥 = Sketch:
𝜎 𝑥 = =2.59
We want to find 𝑃( 𝑥 >193)
𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 193, 1× , 188, 2.59 =0.0268
𝜇 𝑥 =188
𝜎 𝑥 =2.593
𝑁𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓(193, , 188, 2.593) =