Question

Mean: 12.96 mm

Standard Deviation: 0.87 mm

Assuming that these data conform to a normal distribution, calculate the following:

a. What would be the Z-Score for 14.2 mm in length?

b. What percentage of a large set of examples would fall between 11.4mm and 14.8 mm?

c. Assuming that we have access to a sample of 4,593 examples, how many would we expect to fall within the above range?

Answer 1

part a)

Z = ( X - µ ) / σ
Z = ( 14.2 - 12.96 ) / 0.87
Z = 1.43

Part b)

X ~ N ( µ = 12.96 , σ = 0.87 )
P ( 11.4 < X < 14.8 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 11.4 - 12.96 ) / 0.87
Z = -1.7931
Z = ( 14.8 - 12.96 ) / 0.87
Z = 2.1149
P ( -1.79 < Z < 2.11 )
P ( 11.4 < X < 14.8 ) = P ( Z < 2.11 ) - P ( Z < -1.79 )
P ( 11.4 < X < 14.8 ) = 0.9828 - 0.0365
P ( 11.4 < X < 14.8 ) = 0.9463

Percentage = 0.9463 * 100 = 94.63%

Part b)

N = 4593

Expected samples are 4593 * 0.9463 = 4346.36 ≈ 4346 samples fall within the above range.