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 117 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 95?% of bags.

?(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip? cookies?

Answer 1

Here, we are given the distribution as:

a) From standard normal tables we get:

P(Z < -0.6128 ) = 0.27

Therefore the 27th percentile observation here is computed as:

= Mean -0.6128*Std Dev.

= 1261 -0.6128*117

= 1189.3024

b) From standard normal tables, we get:

P( -1.96 < Z <1.96 ) = 0.95

Therefore the interval here is computed as:

Mean - 1.96*Std Dev. , Mean - 1.96*Std Dev.

1261 - 1.96*117, 1261 + 1.96*117

1031.68, 1490.32

This is the required middle interval required here

c) From the standard normal tables, we get:

P(Z < -0.6745 ) = 0.25

Therefore, P(Z > 0.6745) = 0.25

Therefore, here the interval is computed as:

1261 - 0.6745*117, 1261 + 0.6745*117

Therefore interquartile range here is computed as:

= 2*0.6745*117

= 157.833

Therefore the required interquartile range is 157.833