Question

The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with a mean of 1261 chips and a standard deviation of 118 chips.
​(a) Determine the 27th percentile for the number of chocolate chips in a bag.
​(b) Determine the number of chocolate chips in a bag that make up the middle 97​% of bags.
​(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip​ cookies?

Answer 1

Mean, = 1261

Standard deviation, = 118

P(X < A) = P(Z < (A - )/)

a. Let the 27th percentile be T

P(X < T) = 0.27

P(Z < (T - 1261)/118) = 0.27

Take the value of Z corresponding to 0.27 from standard normal distribution table.

(T - 1261)/118 = -0.61

T = 1189.02

b. Let the middle 97% be between M and N

P(X < M) = 0.5 - 0.97/2 = 0.015

P(Z < (M - 1261)/118) = 0.015

(M - 1261)/118 = -2.17

M = 1004.94

P(X < N) = 0.5 + 0.97/2 = 0.985

P(Z < (N - 1261)/118) = 0.985

(N - 1261)/118 = 2.17

N = 1517.06

c. The interquartile range, IQR = Q3 - Q1

P(X < Q3) = 0.75

P(Z < (Q3 - 1261)/118) = 0.75

(Q3 - 1261)/118 = 0.67

Q3 = 1340.06

(Q1 - 1261)/118 = -0.67

Q1 = 1181.94

IQR = 1340.06 - 1181.94

= 158.12