Assume that the annual household expenditure on sugar is approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10. what is the probability that a household spent more than $10.00?

 To solve this problem, we will have to make use of the z statistic. The formula for the z score is given as:

z = (x – u) / s

where,

x is the sample value = more than $10.00

u is the sample mean = $8.22

s is the standard deviation = $1.10

Substituting the values into the equation to solve for z:

z = (10 – 8.22)  / 1.10

z = 1.78 / 1.10

z = 1.62

We then look for the p value using the standard distribution tables at the specified z score value = 1.62. Since this is a right tailed test, therefore the p value is:

p = 0.0526

or

p = 5.26%

Therefore there is a 5.26% probability that a household spent more than $10.00