Using the normal distribution, the indicated probability is given as follows:
P(7 ≤ x ≤ 11) = 0.6247.
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 mean and the standard deviation are given, respectively, by:
[tex]\mu = 8, \sigma = 2[/tex]
P(7 ≤ x ≤ 11) is the p-value of Z when X = 11 subtracted by the p-value of Z when X = 7, hence:
X = 11:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Z = (11 - 8)/2
Z = 1.5
Z = 1.5 has a p-value of 0.9332.
X = 7:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Z = (7- 8)/2
Z = -0.5
Z = -0.5 has a p-value of 0.3085.
0.9332 - 0.3085 = 0.6247.
More can be learned about the normal distribution at https://brainly.com/question/28096232
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