Practice Exercise A. Random samples of size 225 are drawn from a population with mean 100 and standard deviation 20. 1. Find the mean of sampling distribution of the means? 2. Find the standard deviation of sampling distribution of the means? B. The heights of male college students are normally distributed with mean of 68 inches and standard deviation of 3 inches. A random samples of 25 students are drawn with replacement from the population, what would be the (a.) mean, (b) variance and (c) standard deviation of the resulting sampling distribution of the means?
Practice Exercise A. Random samples of size 225 are drawn from a population with mean 100 and standard deviation 20. 1. Find the mean of sampling distribution of the means? 2. Find the standard deviation of sampling distribution of the means? B. The heights of male college students are normally distributed with mean of 68 inches and standard deviation of 3 inches. A random samples of 25 students are drawn with replacement from the population, what would be the (a.) mean, (b) variance and (c) standard deviation of the resulting sampling distribution of the means?
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Question
Please answer the exercise below
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So using the above information we can now solve.
a. The mean of the sampling distribution of the means is equal to the population mean.
→ Since the population mean (μ) is 60,
x= μ
Therefore:
#₁ = 60
b. The variance of the sampling distribution of the means is equal to the population variance
divided by the sample size.
→ Since the population standard deviation () is 5 and the sample size (12) is 16, then:
of = 16
o=1.5625
16
c. The standard deviation of the sampling distribution of the means is equal to the population
standard deviation divided by the square root of the sample size.
→ Since the population standard deviation () is 5 and the sample size (n) is 16, then:
0
5
√16
o₂ = 1.25
0x
Example 2:
The mean and standard deviation of the tax value of all vehicles registered in a certain state are =
P13 525 and = P 4 180. Suppose random samples of size 100 are drawn from the population of vehicles.
What are the (a.) mean, (b.) variance, and (c.) standard deviation of sampling distribution of the means.
Solution:
GIVEN: population mean (μ) =P13 525, population standard deviation (o)=P4 180
sample size (n) = 100
a. The mean of the sampling distribution of the means is equal to the population mean.
→ Since the population mean () is P13 525
Hx = μ
Hy P13 525
Therefore:
b. The variance of the sampling distribution of the means is equal to the population variance
divided by the sample size.
→ Since the population standard deviation () is P4 180 and the sample size (n) is 100, then:
4180²
of
100
0}
17 472 400 P174 724
=
100
c. The standard deviation of the sampling distribution of the means is equal to the population
standard deviation divided by the square root of the sample size.
→ Since the population standard deviation () is P4 180 and the sample size (n) is 100, then:
0₂
@
5
4180
0,
√100
4180
=P418
10
Now it's your turn to practice.
Practice Exercise
A. Random samples of size 225 are drawn from a population with mean 100 and standard deviation 20.
1. Find the mean of sampling distribution of the means?
2. Find the standard deviation of sampling distribution of the means?
B. The heights of male college students are normally distributed with mean of 68 inches and standard
deviation of 3 inches. A random samples of 25 students are drawn with replacement from the
population, what would be the (a.) mean, (b) variance and (c) standard deviation of the resulting
sampling distribution of the means?
Done
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