Suppose the wrapper of a certain candy bar lists its weight as 2.13 ounces. Natu

Question

Suppose the wrapper of a certain candy bar lists its weight as 2.13 ounces. Naturally, the weights of individual bars vary somewhat. Suppose also that the weights of these candy bars vary according to a normal distribution with mean mu equals 2.2 ounces and standard deviation equals 0.04 ounces.

a) Determine the probability that the sample mean weight would be less than 2.15 ounces.

b) Determine the value such that the probability is 0.0125 of obtaining a sample mean weight of less than that value.

Explanation / Answer

(The number of bars n is missing!)

= 2.2 = 0.04

a) X = 2.15

Z = (X - ) /

= (2.15 - 2.2) / 0.04

= -0.05 / 0.04

= -1.25

Probability from tables = 0.1056.

b) A probability = 0.0125 corresponds to a z value of -2.24

=> X = + z

= 2.2 - 2.24*0.04

= 2.1104.

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