Question

A snack food producer sells bags labeled as “12 ounces” of its “Cheesy Chips.” Due to natural variations in the creation of these chips, the weight of Cheesy Chips in each bag is a random variable that follows a normal distribution with a mean of 12.02 ounces and a standard deviation of 0.16 ounces.

1. Find the probability that a randomly selected bag of Cheesy Chips will weigh less than 11.8 ounces. (5 points)

b. A random sample of 15 bags of Cheesy Chips has a mean weight of 11.8 ounces. What is the probability of a random sample of 15 bags of Cheesy Chips averaging less than 11.8 ounces occurring? (5 points)

c. How do your answers differ for parts a and b and why did you get such different answers for a and b?

Answer 1

a) P(x < 11.8)

= P[(x - ) / < (11.8 - 12.02) / 0.16]

= P(z < -1.375)

Using z table,

= 0.0846

b) = 12.02

n = 0.16 / 15

P( < 11.8) = P(( - ) / < (11.8 - 12.02) / 0.16 / 15 )

= P(z < -5.32)

Using z table

= 0

c) A sample size is larger in part (b), so standard deviation is smaller in part (b), so probability is smaller than part (a).