Using the normal distribution, it is found that there is a 0.9938 = 99.38% probability that the average of these members is less than 35.
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]
In this problem, we have that the parameters are given as follows:
[tex]\mu = 34, \sigma = 4, n = 100, s = \frac{4}{\sqrt{100}} = 0.4[/tex].
The probability that the average of these members is less than 35 is the p-value of Z when X = 35, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{35 - 34}{0.4}[/tex]
Z = 2.5
Z = 2.5 has a p-value of 0.9938.
0.9938 = 99.38% probability that the average of these members is less than 35.
More can be learned about the normal distribution at https://brainly.com/question/24663213
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