Question

5. People are selected to serve on juries by randomly picking names from the list of registered voters. The average age for registered voters in the country is μ = 39.7 years with σ = 11.8. The distribution of ages is approximately normal.

During a recent jury trial in the country courthouse, a statistician noted that the average age for the 12 jurors was = 50.4 years.

a. What is the probability of getting a jury with this average age or older by chance.

b. Is it reasonable to conclude that this jury was selected by a random sample of registered voters? Why?

During another recent jury trial in the country courthouse, a statistician noted that the average age for the 12 jurors was = 30.2 years.

c. What is the probability of getting a jury with this average age or younger by chance?

d. Is it reasonable to conclude that this jury was elected by a random sample of registered voters? Why?

e. Find the 90th percentile for the age of an individual jury member in the full population.

f. Find the 90th percentile for the average age of jury members (n = 12).

Answer 1

Normal mu = 39.7 sigma = 11.8 avg age rightarrow 50.4 years Probability of getting a jury with this average age or order by chance p(xgeg50.4) z = xbar-mu/sigma/vn = 50.4-39.7/11.8/sqrt12 z = 0.91sqrt12 = 3.141 p(x geg 50.4 > = p(z > 3.14) = 1-p(z < 3.41) = 0.0001 It is equal to 0.01% which is very small left is not responsible avg age = 30.2 year n = 12 p(x c 30.2) = p(z < 30.2-39.7/11.8/sqrt12) = p(z < -2.78) probability = 0.0026 = 0.26%\r\n mu = 39.7 sigma = 11.8 90^th percentile alpha = 0.1 from z table p(z < 1.28) = 0.9 It means 90^th percentile corresponds to z alpha = 1.28 z = x-mu/sigma x = mu + z sigma = 39.7 + 1.28(11.8) x = 54.8 90^th percentile rightarrow 54.8 years for Jury members n = 12 X = mu + z sigma / squareroot n = 39.7 + 1.28(11.8/sqrt12) = 39.7 + 4.36 = 44.06 90^th percentile for Jury members = 44.06 years