Question

Based on the model ​N(1156​,85​) describing steer​ weights, what are the cutoff values for ​a) the highest​ 10% of the​ weights? ​b) the lowest​ 20% of the​ weights? ​c) the middle​ 40% of the​ weights?

Answer 1

Solution :

Given that,  

mean = = 1156

standard deviation = = 85

X N (1156 , 85)

a)

Using standard normal table ,

P(Z > z) = 10%

1 - P(Z < z) = 0.1

P(Z < z) = 1 - 0.1

P(Z < 1.28) = 0.9

z = 1.28

Using z-score formula,

x = z * +

x = 1.28 * 85 + 1156 = 1264.8

weights = 1264.8

b)

Using standard normal table ,

P(Z < z) = 20%

P(Z < -0.84) = 0.2

z = -0.84

Using z-score formula,

x = z * +

x = -0.84 * 85 + 1156 = 1084.6

weights = 1084.6

c)

Middle 40% as the to z values are -0.524 and 0.524

Using z-score formula,

x = z * +

x = -0.524 * 85 + 1156 = 1111.46

and

x = 0.524 * 85 + 1156 = 1200.54

the values are 1111.46 and 1200.54