Answer to Question #225983 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 is normally distributed. What is the probability of a randomly chosen family having an income less than $36 000? Provide evidence of your work by typing out your full solution.

Answer 1

$P(X>50000) = 1-P(X≤50000) = 1- 0.3085 = 0.6915$

Probability of a randomly chosen family having an income greater than $50 000

$P(X>50000) = 0.6915$

Probability that a family income is less than $36 000

$P(X≤36000)$

Z-score for 36000

$=\frac{(36000-60000)}{20000 }\\ =\frac{ -24000}{20000}\\ = -1.2$

From standard normal tables , $P(Z≤ -1.2) =0.1151$