Respuesta :
Using the normal distribution, there is a 0.2148 = 21.48% probability that the sum of the 40 values is less than 7,100.
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].
For this problem, these parameters are given as follows:
[tex]\mu = 180, \sigma = 20, n = 40, s = \frac{20}{\sqrt{40}} = 3.1623[/tex]
A sum of 7100 is equivalent to a sample mean of 7100/40 = 177.5, which means that the probability is the p-value of Z when X = 177.5, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{177.5 - 180}{3.1623}[/tex]
Z = -0.79
Z = -0.79 has a p-value of 0.2148.
There is a 0.2148 = 21.48% probability that the sum of the 40 values is less than 7,100.
More can be learned about the normal distribution at https://brainly.com/question/28135235
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