Using the normal distribution, it is found that the mean is of [tex]\mu = 37.27[/tex] and the standard deviation is of [tex]\sigma = 5.68[/tex].
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 p-value of Z when X = 30 is of 0.1, hence, when X = 30, Z = -1.28, so:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.28 = \frac{30 - \mu}{\sigma}[/tex]
[tex]30 - \mu = -1.28\sigma[/tex]
[tex]\mu = 30 + 1.28\sigma[/tex]
The p-value of Z when X = 32.5 is of 0.2, hence when X = 32.5, Z = -0.84, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.84 = \frac{32.5 - \mu}{\sigma}[/tex]
[tex]32.5 - \mu = -0.84\sigma[/tex]
[tex]\mu = 32.5 + 0.84\sigma[/tex]
Hence:
[tex]30 + 1.28\sigma = 32.5 + 0.84\sigma[/tex]
[tex]0.44\sigma = 2.5[/tex]
[tex]\sigma = \frac{2.5}{0.44}[/tex]
[tex]\sigma = 5.68[/tex]
[tex]\mu = 32.5 + 0.84(5.68) = 37.27[/tex]
More can be learned about the normal distribution at https://brainly.com/question/27643290
#SPJ1
