Question

i. Family incomes have a mean of $60,000 with a standard deviation of $20,000. The data are normally distributed. What is the probability of a randomly chosen family having an income greater than $50 000?

ii. On average, weight of carry-on baggage of passengers on planes is 32.2 pounds. Assuming a standard deviation of 4.3 pounds, find the probability that the average weight of carry-on baggage of a random sample of 40 passengers exceeds 30 pounds.

iii. Suppose you are working with a data set that is normally distributed, with a mean of 150 and a standard deviation of 46. Determine the value of x from the following information. (Round your answers and z values to 2 decimal places.)

a. 60% of the values are greater than x. b. x is less than 16% of the values.

c. 25% of the values are less than x.

d. x is greater than 65% of the values.

Answer 1

i. We have that

$\mu=60000$

$\sigma=20000$

$P(X>50000)=1-P(X<50000)=1-P(Z<\frac{x-\mu}{\sigma})=$

$=1-P(Z<\frac{50000-60000}{20000})=1-P(Z<-0.5)=1-0.3085=0.6915$

ii. We have that

$\mu=32.2$

$\sigma=4.3$

$n=40$

$P(X>30)=1-P(X<30)=1-P(Z<\frac{x-\mu}{\frac{\sigma}{\sqrt n}})=$

$=1-P(Z<\frac{30-32.2}{\frac{4.3}{\sqrt{40}}})=1-P(Z<-3.24)=1-0.0006=0.9994$

iii. We have that

$\mu=150$

$\sigma=46$

a) $P(X>x) =0.6\implies P(X<x)=1-0.6=0.4 \implies Z = -0.25$

$\frac{x-\mu}{\sigma}=\frac{x-150}{46}=-0.25 \implies x=138.5$

b) $P(X<x)=0.16 \implies Z=-0.99 \implies \frac{x-150}{46}=-0.99 \implies x=104.46$

c) $P(X<x)=0.25 \implies Z=-0.67 \implies \frac{x-150}{46}=-0.67 \implies x=119.18$

d) $P(X>x) =0.65\implies P(X<x)=1-0.65=0.35 \implies Z = -0.38$

$\frac{x-150}{46}=-0.38 \implies x=132.52$

Answer:

i. 0.6915

ii. 0.9994

iii. a) 138.5 b) 104.46 c) 119.18 d) 132.52