A normal population has a mean of 19 and a standard deviation of 5.(a) Compute the z value associated with 25 (Round your answer to 2 decimal places.)(b) What proportion of the population is between 19 and 25? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)(c) What proportion of the population is less than 17? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

Answer:a) [tex]Z = 1.2[/tex]b) 38.49% of the population is between 19 and 25.c) 34.46% of the population is less than 17.Step-by-step explanation:Problems of normally distributed samples can be solved using the z-score formula.In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by[tex]Z = \frac{X - \mu}{\sigma}[/tex]After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. If we need to find the probability that the measure is larger than X, it is 1 subtracted by this pvalue.For this problem, we have thatA normal population has a mean of 19 and a standard deviation of 5, so [tex]\mu = 19, \sigma = 5[/tex].(a) Compute the z value associated with 25This is Z when [tex]X = 25[/tex][tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{25 - 19}{5}[/tex][tex]Z = 1.2[/tex](b) What proportion of the population is between 19 and 25? This is the pvalue of Z when [tex]X = 25[/tex] subtracted by the pvalue of Z when [tex]X = 19[/tex].X = 25 has [tex]Z = 1.2[/tex], that has a pvalue of 0.8849.X = 19 has [tex]Z = 0[/tex], that has a pvalue of 0.5000.So 0.8849-0.500 = 0.3849 = 38.49% of the population is between 19 and 25.(c) What proportion of the population is less than 17?This is the pvalue of Z when [tex]X = 17[/tex][tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{17 - 19}{5}[/tex][tex]Z = -0.40[/tex][tex]Z = -0.40[/tex] has a pvalue of 0.3446.This means that 34.46% of the population is less than 17.