Question

The number of chocolate chips in an 18-ounce bag of Chips Ahoy! chocolate chip cookies is approximately normally distributed with a mean of 1262 chops and standard deviation 118 chips according to a study by cadets of U.S. Air Force Academy. (Source: Brad Warner and Jim Rutledge, Chance 12(1): 10-14, 1999)

(a) What is the probability that a randomly selected 18-ounce bag of Chips Ahoy! contains fewer than 1000 chocolate chips? (use 2-table)

(b) What is the probability that a randomly selected 18-ounce bag of Chips Ahoy! contains more than 1200 chocolate chips?

(c) What is the probability that a randomly selected 18-ounce bag of Chips Ahoy! contains between 1000 and 1400 chocolate chips, inclusive?

Answer 1

a) 0.013

b) 0.7

c) 0.866

Explanation:

We are given the following information in the question:

Mean, μ = 1262

Standard Deviation, σ = 118

We are given that the distribution of number of chocolate chips is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

a) P(fewer than 1000 chocolate chips)

P(x < 1000)

`P( x < 1000) = P( z < \displaystyle(1000 - 1262)/(118)) = P(z < -2.220)`

Calculation the value from standard normal z table, we have,

`P(x < 1000) = 0.013 = 1.3\%`

b) P(contains more than 1200 chocolate chips)

`P(x > 1200) = P(z > \displaystyle(1200-1262)/(118)) = P(z > -0.5254)\\\\P( z > -0.5254) = 1 - P(z \leq -0.5254)`

Calculating the value from the standard normal table we have,

`1 - 0.300 = 0.7 = 70\%\\P( x > 1200) = 0.7`

c)P(contains between 1000 and 1400 chocolate chips)

`P(1000 \leq x \leq 1400) = P(\displaystyle(1000 - 1262)/(118) \leq z \leq \displaystyle(1400-1262)/(118)) = P(-2.22 \leq z \leq 1.169)\\\\= P(z \leq 1.169) - P(z < -2.22)\\= 0.879 - 0.013 = 0.866 = 86.6\%`

`P(1000 \leq x \leq 1400) = 86.6\%`