Question

Suppose that the amount of sugar in a can of Pepsi is normally distributed with mean μ = 55 g
and standard deviation σ = 1.5 g. What is the probability that

d. What is the amount of sugar such that we can know that 97.5% of the cans will have less than that
amount of sugar?
e. Suppose that at the plant we randomly sample 25 cans. What is the probability that their average
amount of sugar is more than 56 grams?

Answer 1

Let X : the amount of sugar in a can of Pepsi

Given, X follows normal distribution with mean = 55 and standard deviation = 1.5, i.e., X ~ N(=55, =1.5)

d.) To find the amount of sugar such that we can know that 97.5% of the cans will have less than that amount of sugar, i.e., P(X<x) = 0.975

Using Central Limit Theorem, which states that,

Hence,

Now, we look for the value of (x-55)/1.5 in the standard normal probability table for which the probability is 0.975

We see that P(Z<1.96)=0.975

Thus,

Hence, the the amount of sugar such that we can know that 97.5% of the cans will have less than that amount of sugar is 57.94

e.) To find the probability that the average amount of sugar in 25 randomly chosen sample of sugar can is more than 56 grams.

We know that,

Here, n=25. Hence,

Now, to find   , using CLT,

From z-score table,

Thus,

Therefore, the probability that the average amount of sugar in 25 randomly chosen sample of sugar can is more than 56 grams is 0.0004