Question

4. The number of chocolate chips in an 18-ounce bag of Chips Ahoy! cookies is approximately normally distributed with a mean of µ = 1262 chips and standard deviation σ = 188 chips. Source: Brad Warner and Jim Rutledge, Chance, 12(1): 10-14, 1999

(a) (3 points) What is the probability that a randomly selected 18-ounce bag of Chips Ahoy! contains between 1000 and 1400 chocolate chips, inclusive? Round your answer to 4 decimal places. Problem 4 continued:

(b) (3 points) What proportion of 18-ounce bags of Chips Ahoy! contain more than 1200 chocolate chips? Round your answer to 4 decimal places.

(c) (3 points) What is the percentile of an 18-ounce bag of Chips Ahoy! that contains 1068 chocolate chips?

(d) (3 points) What is the range of chocolate chip counts that accounts for the middle 90% of all 18-ounce bag of Chips Ahoy?

e) (3 points) Suppose you take a sample of 100 bags of Chips Ahoy! cookies, and you find the sample mean number of chips for the 100 bags. What is the probability that the sample mean is greater than 1200? Round your answer to 4 decimal places.

Answer 1

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	1262
std deviation   =σ=	188.0000

probability =

P(1000<X<1400)

=

P(-1.39<Z<0.73)=

0.7673-0.0823=

0.6850

b)

probability =

P(X>1200)

=

P(Z>-0.33)=

1-P(Z<-0.33)=

1-0.3707=

0.6293

c)

probability =

P(X<1068)

=

P(Z<-1.03)=

0.1515 ~15.15th percentile

d)

for middle 90% ; critical z = -/+1.645

hence corresponding range =mean -/+ z*std error =1262 -/+ 1.645*188 =952.74 to 1571.26

e)

sample size       =n=	100
std error=σx̅=σ/√n=	18.8000

probability =

P(X>1200)

=

P(Z>-3.3)=

1-P(Z<-3.3)=

1-0.0005=

0.9995