Using the normal distribution, it is found that the probabilities are given as follows:
a) 0.8871 = 88.71%.
b) 0.0778 = 7.78%.
c) 0.8485 = 84.85%.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation
.
The parameters in this problem are given as follows:
![\mu = 1140, \sigma = 310, n = 16, s = \frac{310}{\sqrt{16}} = 77.5](https://tex.z-dn.net/?f=%5Cmu%20%3D%201140%2C%20%5Csigma%20%3D%20310%2C%20n%20%3D%2016%2C%20s%20%3D%20%5Cfrac%7B310%7D%7B%5Csqrt%7B16%7D%7D%20%3D%2077.5)
Item a:
The probability is the <u>p-value of Z when X = 1250 subtracted by the p-value of Z when X = 1000</u>, hence:
X = 1250:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{1250 - 1140}{77.5}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B1250%20-%201140%7D%7B77.5%7D)
Z = 1.42
Z = 1.42 has a p-value of 0.9222.
X = 1000:
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{1000 - 1140}{77.5}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B1000%20-%201140%7D%7B77.5%7D)
Z = -1.81
Z = -1.81 has a p-value of 0.0351.
0.9222 - 0.0351 = 0.8871 = 88.71% probability.
Item b:
The probability is <u>one subtracted by the p-value of Z when X = 1250</u>, hence:
1 - 0.9222 = 0.0778 = 7.78%.
Item c:
The probability is the <u>p-value of Z when X = 1220</u>, hence:
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{1220 - 1140}{77.5}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B1220%20-%201140%7D%7B77.5%7D)
Z = 1.03
Z = 1.03 has a p-value of 0.8485.
0.8485 = 84.85% probability.
More can be learned about the normal distribution at brainly.com/question/4079902
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