Question

The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with a mean of 1262 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 98​% of bags. ​

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

Answer 1

a) P(Z < z) indicates a percentile
27th percentile implies P(Z < z) = 0.2700
The z value which separates the lower 0.2700 area from the rest is - 0.61 approximately
z = (X-Mean)/SD
X = Mean + (z*SD)
Number of chocolate chips which represent 27th percentile = X = 1262 + (- 0.61*118)
= 1190 on rounding as needed

b) Middle 96% leaves 4% area equally in the left and right tails of the normal curve
Area in the left tail = Area in the right tail = 4%/2 = 2% = 0.0200
The z value which separates the bottom 0.0200 area from the rest is - 2.05 approximately
The z value which separates the top 0.0200 area from the rest is + 2.05 approximately
X1 = 1262 + (- 2.05*118) = 1020
X2 = 1262 + (2.05*118) = 1503
Required interval of the chocolate chips is from 1020 to 1503

c) Interquartile range

The z value which separates the bottom 0.25 area from the rest is - 0.6745 approximately
The z value which separates the top 0.75 area from the rest is + 0.6745 approximately

X1 = 1262 + (- 0.6745*118) = 1182
X2 = 1262 + (0.6745*118) = 1341

Interquartile range = P(0.25 < P < 0.75) = 1341 - 1182

= 159