MNM Corporation gives each of its employees an aptitude test. The scores on the test are normally distributed with a mean of 75 and a standard deviation of 15. A simple random sample of 25 is taken from a population of 500.
a. What are the expected value, the standard deviation, and the shape of the sampling distribution of?
b. What is the probability that the average aptitude test in the sample will be between 70.14 and 82.14?

c. What is the probability that the average aptitude test in the sample will be greater than 82.68?
d. What is the probability that the average aptitude test in the sample will be less than 78.69?
e. Find a value, C, such that P(( x>C) = .015.

Answer:

a) The expected value is 75, the standard deviation is 3 and the shape is approximately normal.

b) 0.9387 = 93.87% probability that the average aptitude test in the sample will be between 70.14 and 82.14.

c) 0.0052 = 0.52% probability that the average aptitude test in the sample will be greater than 82.68.

d) 0.8907 = 89.07% probability that the average aptitude test in the sample will be less than 78.69.

e) C = 81.51.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

The scores on the test are normally distributed with a mean of 75 and a standard deviation of 15.

This means that [tex]\mu = 75, \sigma = 15[/tex]

a. What are the expected value, the standard deviation, and the shape of the sampling distribution of?

By the Central Limit Theorem, it will be approximately normal, with expected value [tex]\mu = 75[/tex] and standard deviation [tex]s = \frac{15}{\sqrt{25}} = 3[/tex].

b. What is the probability that the average aptitude test in the sample will be between 70.14 and 82.14?

This is the pvalue of Z when X = 82.14 subtracted by the pvalue of Z when X = 70.14.

X = 82.14

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{82.14 - 75}{3}[/tex]

[tex]Z = 2.38[/tex]

[tex]Z = 2.38[/tex] has a pvalue of 0.9913.

X = 70.14

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{70.14 - 75}{3}[/tex]

[tex]Z = -1.62[/tex]

[tex]Z = -1.62[/tex] has a pvalue of 0.0526

0.9913 - 0.0526 = 0.9387

0.9387 = 93.87% probability that the average aptitude test in the sample will be between 70.14 and 82.14.

c. What is the probability that the average aptitude test in the sample will be greater than 82.68?

This is 1 subtracted by the pvalue of Z when X = 82.68. So

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{82.68 - 75}{3}[/tex]

[tex]Z = 2.56[/tex]

[tex]Z = 2.56[/tex] has a pvalue of 0.9948.

1 - 0.9948 = 0.0052

0.0052 = 0.52% probability that the average aptitude test in the sample will be greater than 82.68.

d. What is the probability that the average aptitude test in the sample will be less than 78.69?

This is the pvalue of Z when X = 78.69. So

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{78.69 - 75}{3}[/tex]

[tex]Z = 1.23[/tex]

[tex]Z = 1.23[/tex] has a pvalue of 0.8907.

0.8907 = 89.07% probability that the average aptitude test in the sample will be less than 78.69.

e. Find a value, C, such that P(( x>C) = .015.

This is X when Z has a pvalue of 1 - 0.015 = 0.985. So X when Z = 2.17.

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]2.17 = \frac{X - 75}{3}[/tex]

[tex]X - 75 = 3*2.17[/tex]

[tex]X = 81.51[/tex]

So C = 81.51.

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