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) The value of C = 81.51.
What is meant by 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 z-score of a measure X is given by:
[tex]$Z=\frac{X-\mu}{\sigma}$[/tex]
The Z-score calculates the deviation of the measure from the mean in standard deviations. We glance at the z-score table after determining the Z-score to determine the p-value connected to it. The likelihood that the measure's value is less than X, or the percentile of X, is represented by this p-value. The likelihood that the value of the measure is greater than X is obtained by deducting 1 from the p-value.
The Central Limit Theorem establishes 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. 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. The p-value of Z when X = 82.14 subtracted by the p-value 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]
Z = 2.38
Z = 2.38 has a p-value of 0.9913.
[tex]$Z=\frac{X-\mu}{s}$[/tex]
substitute the values in the above equation, we get
[tex]$Z=\frac{70.14-75}{3}$[/tex]
Z = -1.62 has a p-value 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. This is 1 subtracted by the p-value of Z when X=82.68.
[tex]$Z=\frac{X-\mu}{s}[/tex]
substitute the values in the above equation, we get
[tex]${data-answer}amp;Z=\frac{82.68-75}{3} \\[/tex]
Z = 2.56 has a p-value 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. This is the p-value of Z when X=78.69. So
[tex]${data-answer}amp;Z=\frac{X-\mu}{s} \\[/tex]
substitute the values in the above equation, we get
[tex]${data-answer}amp;Z=\frac{78.69-75}{3} \\[/tex]
Z = 1.23 has a p-value 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) = 0.015.
This is X when Z has a p-value of 1 - 0.015 = 0.985.
So X when Z = 2.17.
[tex]$Z=\frac{X-\mu}{s}$[/tex]
substitute the values in the above equation, we get
[tex]$2.17=\frac{X-75}{3}$[/tex]
X - 75 = 3 × 2.17
X = 81.51
Therefore, the value of C = 81.51
The complete question is:
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.
To learn more about Normal probability distribution refer to:
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