Respuesta :
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%.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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 [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
The parameters in this problem are given as follows:
[tex]\mu = 1140, \sigma = 310, n = 16, s = \frac{310}{\sqrt{16}} = 77.5[/tex]
Item a:
The probability is the p-value of Z when X = 1250 subtracted by the p-value of Z when X = 1000, hence:
X = 1250:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1250 - 1140}{77.5}[/tex]
Z = 1.42
Z = 1.42 has a p-value of 0.9222.
X = 1000:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1000 - 1140}{77.5}[/tex]
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 one subtracted by the p-value of Z when X = 1250, hence:
1 - 0.9222 = 0.0778 = 7.78%.
Item c:
The probability is the p-value of Z when X = 1220, hence:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1220 - 1140}{77.5}[/tex]
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 https://brainly.com/question/4079902
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