Using the normal distribution, there is a 0.4826 = 48.26% probability that the sample mean is between 15 and 16 grams per day.
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]
For this problem, the parameters are given as follows:
[tex]\mu = 15, \sigma = 3, n = 40, s = \frac{3}{\sqrt{40}} = 0.4743[/tex]
The probability is the p-value of Z when X = 16 subtracted by the p-value of Z when X = 15, hence:
X = 16:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{16 - 15}{0.4743}[/tex]
Z = 2.11
Z = 2.11 has a p-value of 0.9826.
X = 15:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{15 - 15}{0.4743}[/tex]
Z = 0
Z = 0 has a p-value of 0.5.
0.9826 - 0.5 = 0.4826 = 48.26% probability that the sample mean is between 15 and 16 grams per day.
More can be learned about the normal distribution at https://brainly.com/question/15181104
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