The weights of packages of cheese from a dairy are approximately Normally distributed. Based on a very
large sample, it was found that 15 percent of the
packages weighed less than 15.48 ounces, and 10
percent weighed more than 16.64 ounces.
What are the mean and standard deviation of the
weights of the packages of cheese?
OZ
U=
0 =
OZ

U = 16 oz

0 = 0.5 oz

Step-by-step explanation:

Use z-table to find z-scores for the percents

z-score for 15% = -1.04

z-score for 90% = 1.28

z = x - U / 0

First equation: -1.04 = 15.48 - U / 0

Second equation: 1.28 = 16.64 - U / 0

Solve first equation for U: U = 1.04 0 + 15.48

Multiply both sides of the second equation by 0: 1.28 0 = 16.64 - U

Substitute the first equation for U into the second equation and solve for 0: 1.28 0 = 16.64 - (1.04 0 + 15.48)

1.28 0 = 16.64 - 1.04 0 - 15.48

2.32 0 = 1.16

0 = 0.5

Substitute the value of 0, 0.5, into the first equation and then solve for U: 1.04 (0.5) + 15.48 = 16

the answer is y = 11.54

hope this : )