Question

During the past six months, 73.2 percent of U.S. households purchased sugar. Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10. Find the probability that a household spent less than $5.00 on sugar.

Answer 1

Answer:

0.0017 is the probability that a household spent less than $5.00 on sugar.

Step-by-step explanation:

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $8.22

Standard Deviation, σ =  $1.10

We are given that the distribution of expenditure is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P( household spent less than $5.00 on sugar)

P(x < 5)

$P( x< 5) = P( z < \displaystyle\frac{5 - 8.22}{1.10}) = P(z < -2.9272)$

Calculation the value from standard normal z table, we have,  

$P(x < 5) = 0.0017$

0.0017 is the probability that a household spent less than $5.00 on sugar.