A candy maker produces mints that have a label weight of 20.4 grams. Assume that the distribution of the weights of these mints is N(21.37, 0.16).

(a) Let X denote the weight of a single mint selected at random from the production line. Find P(X > 22.07).

(b) Suppose that 15 mints are selected independently and weighed. Let Y equal the number of these mints that weigh less than 20.857 grams. Find P(Y ≤ 2).

Step 1 of 3:

Given that a candy maker produces mints having weight 20.4 grams.

Also it is given that the weight of the mints have a normal distribution with parameters 21.37 and 0.16.

That is weight of the mints have N(21.37,0.16) distribution.

Thus mean of the weight of the mints,=21.37 and variance of the weights of the mints,

=0.16.

Step 2 of 3:

(a)

Given that let X denotes the weight of a mint selected at random from the production line. We have to find the probability that weight of the mint selected is greater than 22.07 grams.

That we have to find the value of P(X>22.07).

We make use of standard normal approximation. That is Z=~N(0,1).

Thus,

P(X>22.07)=P(>)

=P(Z>)

=P(Z>1.75)

=1-P(Z<1.75)

=1-0.95994

=0.04006

Using the standard normal distribution table we get the value of P(Z<1.75) as 0.95994.This value corresponds to the row containing 1.7 and column containing 0.05.

Thus,P(Z>22.07)=0.04006.