Question

The distribution of weights on 9-ounce bags of potato chips is approximately normal with a mean of 9.12 ounces and a standard deviation of 0.15 ounce. What is the range of weights for 95% of the bags?

Answer 1

Let X be the weight of bags of potato chips

$X \sim N(9.12,0.15)$

Then

$z= \frac{X-9.12}{0.15}\sim N(0,1)$

From z table we find that

For z = -1.96 , X = -1.96*0.15 + 9.12 = 8.826

For z = 1.96, X = 1.96 * 0.15 + 9.12 = 9.414

95% of the bags is between 8.83 ounce and 9.41 ounce