Question

The distribution of the weights of loaves of bread from a certain bakery follows approximately a normal distribution. Based on a very large sample, it was found that 10 percent of the loaves weighed less than 15.34 ounces, and 20 percent of the loaves weighed more than 16.31 ounces. What are the mean and standard deviation of the distribution of the weights of the loaves of bread?A. μ = 15.82, σ =.48
B. μ = 15.82, σ =.69
C. μ = 15.87, σ =.50
D. μ = 15.93, σ =.46
E. μ = 16.00, σ =.50

Answer 1

Answer:

D. μ = 15.93, σ =.46

Step-by-step explanation:

Z = (X –µ)/σ

Let be

X = Loaves weight

µ = mean

σ = standard deviation

let be

X1 = 15.34

X2 = 16.31

We know that 10 percent of the loaves weighed less than X1 and 20 percent of the loaves weighed more than X2. Using a Z table, we find z value for the areas given

If the 10% of the loaves weighed less than X1, 90% weighed more than X1. Z value is -1.282.

If the 20% of the loaves weighed more that X2, 80% weighed less than X2. Z value is 0.842.

Now we have two equations to find µ and σ

-1.282 = (15.34 –µ)/σ  ….. equation 1

.842 = (16.31 –µ)/σ   ….. equation 2

Dividing equation 1 and equation 2

-1.282/0.842 = (15.34 –µ)/ (16.31 –µ)

-1.522 = (15.34 –µ)/ (16.31 –µ)

-1.522 (16.31 –µ) = (15.34 –µ)  

-24.833 + 1.522 µ =15.34 - µ

1.522 µ + µ =15.34 + 24.833

2.522 µ = 40.173

µ = 15.93

Replacing µ in equation 1

σ = 0.46