Question

The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean of 1261 and a standard deviation of 117

​(a) Determine the 26th 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 96%

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

Answer 1

Solution :

mean = = 1261

standard deviation = = 117

Using standard normal table,

a ) P(Z < z) = 26%

P(Z < z) = 0.26

P(Z < -0.6433) = 0.26

z = -0.64

Using z-score formula,

x = z * +

x = - 0.64 * 117 + 1261

= 1186.12

P₂₆ = 1186.12

b ) P(-z < Z < z) = 96%
P(Z < z) - P(Z < z) = 0.96
2P(Z < z) - 1 = 0.96
2P(Z < z ) = 1 + 0.96
2P(Z < z) = 1.96
P(Z < z) = 1.96 / 2
P(Z < z) = 0.98
z = 2.05 znd z = - 2.05

Using z-score formula,

x = z * +

x = 2.05 * 117 + 1261

= 1500.85

x = 1500.85

x = z * +

x = - 2.05 * 117 + 1261

= 1009.45

x = 1009.45

c )  P(Z < z) = 75%

P(Z < z) = 0.75

P(Z < 0.6745) = 0.75

z = 0.67

Using z-score formula,

x = z * +

x = 0.67 * 117 + 1261

   Q₃ = 1339.39

x = z * +

x = -0.67 * 117 + 1261

_{​​​​​​​} Q₁ = 1182.61

The interquartile range = Q₃ - Q₁

₌ 1339.39 - 1182.61

= 156.78

The interquartile range =156.78