Question

The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean

1263

and a standard deviation of

118.

​(a) Determine the

29th

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

solution:-

given that mean = 1263 , standard deviation = 118

(a) the 29th percentile for the number of chocolate chips in a bag

the z value that corresponds to the probability from standard normal table of 0.29 is

z = -0.55

formula

=> mean - z*standard deviation

=> 1263 - 0.55*118

=> 1199

(b) the number of chocolate chips in a bag that make up the middle 97% of bags

from standard normal table

middle 97% has z values are +/- 2.17

=> mean +/- z*standard deviation

=> 1263 +/- 2.17 * 118

=> 1007 to 1519

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

here interquartile range = q3-q1

here first we find q3

here p = 0.75

the z value corresponds to 0.75 is z = 0.67

=> formula

=> mean + z*standard deviation

=> 1263 + 0.67*118

=> 1342

here we find q1

here p = 0.25

the z value corresponds to 0.25 is z = -0.67

=> formula

=> mean - z*standard deviation

=> 1263 - 0.67*118

=> 1184

the interquartile range = q3-q1 = 1342-1184 = 158