Question

The number of chocolate chips in an​ 18-ounce bag of chocolate chip cookies is approximately normally distributed with a mean of chips and standard deviation chips. ​

Required:
a. What is the probability that a randomly selected bag contains between and chocolate​ chips, inclusive? ​
b. What is the probability that a randomly selected bag contains fewer than chocolate​ chips? ​
c. What proportion of bags contains more than chocolate​ chips? ​
d. What is the percentile rank of a bag that contains chocolate​ chips?

Answer 1

Explained below.

Explanation:

The number of chocolate chips in an​ 18-ounce bag of chocolate chip cookies is approximately normally distributed with a mean of 1262 chips and standard deviation 118 chips. ​

(a)

Compute the probability that a randomly selected bag contains between 1000 and 1400 chocolate​ chips as follows:

`P(1000<X<1400)=P((1000-1262)/(118)<(X-\mu)/(\sigma)<(1400-1262)/(118))\\\\=P(-2.22<Z<1.17)\\\\=P(Z<1.17)-P(Z<-2.22)\\\\=0.87900-0.01321\\\\=0.86579\\\\\approx 0.8658`

Thus, the probability that a randomly selected bag contains between 1000 and 1400 chocolate​ chips is 0.8658.

(b)

Compute the probability that a randomly selected bag contains fewer than 1000 chocolate​ chip as follows:

`P(X<1000)=P((X-\mu)/(\sigma)<(1000-1262)/(118))\\\\=P(Z<-2.22)\\\\=0.01321\\\\\approx 0.0132`

Thus, the probability that a randomly selected bag contains fewer than 1000 chocolate​ chip is 0.0132.

(c)

Compute the proportion of bags that contains more than 1200 chocolate​ chips as follows:

`P(X>1200)=P((X-\mu)/(\sigma)>(1200-1262)/(118))\\\\=P(Z<-0.53)\\\\=0.29806\\\\\approx 0.2981`

Thus, the proportion of bags that contains more than 1200 chocolate​ chips is 0.2981.

(d)

Compute the percentile rank of a bag that contains 1125 chocolate​ chips as follows:

`P(X<1125)=P((X-\mu)/(\sigma)<(1125-1262)/(118))\\\\=P(Z<-1.16)\\\\=0.12302\\\\\approx 0.123`

Thus, the percentile rank of a bag that contains 1125 chocolate​ chips is 12.3rd.