Question

The test scores of a population of 10 students are as follow: 29, 35, 39, 42, 44, 45, 47, 50, 53, 59.

Upon looking at how the distribution of these 10 numbers are very close to being symmetric and almost resemble a bell shape, we conclude that the test score distribution approximately follows the normal distribution. And since our samples are drawn from a normally distributed population, we will correctly conclude that the sampling distribution of Xbar is normally distributed too, even though our sample size (n=4) is rather small. What is the average of the sampling distribution of Xbar, (i.e. the Miu of Xbar)? A. 8.283 B. 8.751 C. 9.028 D. 9.754

What is the standard deviation of the sampling distribution of the average (i.e. the sigma Xbar)? Hint: In this question and the questions that follow ignore the fact that n/N is more than 5% and still go ahead and use the simple formula for the standard deviation of Xbar (i.e. Standard Deviation / Square root of n) A. 8.283 B. 8.751 C. 4.142 D. 4.376 E. 2.071

If you randomly select a sample of size 4, what is the probability that the average test score in this sample be between 42 and 46? A. 0 B. 0.15 C. 0.29 D. 0.37 E. 0.66

Answer 1

Solution:

First we have to find the population mean and population standard deviation for the given data.

Mean = µ = ∑X/n

Var = σ² = ∑(X - mean)^2/n

SD = σ = sqrt(Var)

Calculation table is given as below:

No.	X	(X - mean)^2
1	29	234.09
2	35	86.49
3	39	28.09
4	42	5.29
5	44	0.09
6	45	0.49
7	47	7.29
8	50	32.49
9	53	75.69
10	59	216.09
Total	443	686.1
Mean	44.3

Mean = µ = ∑X/n = 443/10 = 44.3

Var = σ² = ∑(X - mean)^2/n

Var = σ² = 686.1/10 = 68.61

SD = σ = sqrt(Var)

SD = σ = sqrt(68.61)

SD = σ = 8.283115356

What is the average of the sampling distribution of Xbar?

We know that the average of the sampling distribution of Xbar is equal to the population mean.

µ_x̄ = µ = 44.3

Answer: 44.3

What is the standard deviation of the sampling distribution of the average?

σ_x̄ = σ/sqrt(n) = 8.283115356/sqrt(4) = 4.14155768

Answer: C. 4.142

If you randomly select a sample of size 4, what is the probability that the average test score in this sample be between 42 and 46?

Here, we have to find P(42<Xbar<46)

P(42<Xbar<46) = P(Xbar<46) – P(Xbar<42)

Find P(Xbar<46)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (46 – 44.3)/[ 8.283115356/sqrt(4)]

Z = 1.7/4.14155768

Z = 0.410474

P(Z<0.410474) = P(Xbar<46) = 0.659271

(by using z-table)

Now, find P(Xbar<42)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (42 – 44.3)/[ 8.283115356/sqrt(4)]

Z = -2.3/4.14155768

Z = -0.55535

P(Z<-0.55535) = P(Xbar<42) = 0.289329

(by using z-table)

P(42<Xbar<46) = P(Xbar<46) – P(Xbar<42)

P(42<Xbar<46) = 0.659271 - 0.289329

P(42<Xbar<46) = 0.369942

P(42<Xbar<46) = 0.37

Answer: D. 0.37