Answer to Question #222914 in Statistics and Probability for Mustafa Zaidi

Question

Family incomes have a mean of $60 000 with a standard deviation of $20 000. The data are normally distributed. Determine the probability that a randomly selected family has a family income less than $36 000. Include evidence of your work by typing out your solution fully.

Answer 1

Solution:

Given:

$\mu=60000$

$\sigma=20000$

$P(X<36000)=P(Z<\frac{36000-60000}{20000})$

$=P(Z<-1.2)=P(Z>1.2)=1-P(Z\le1.2) \\=1-0.88493 \\=0.11507$