A manufacturer of cell phone cases monitors the machines closely to make sure they are producing cases that do not vary far from the desired dimensions. One specific dimension of interest is the width of
the case. The width of produced cases varies Normally with a mean of 2.98 inches and a standard
deviation of 0.05 inches.
a) What proportion of cases produced have a width that is less than 3.014 inches?
b) What is the 25th percentile for the width of the cases produced, in inches?
c) An outlier is any value that falls at least 1.5 times the interquartile range (IQR) above Q3 or below Q1. If the width of a case is an outlier, that case must be discarded. What proportion of all cases
will be discarded?

Step-by-step explanation:

Given the following :

Mean ( m) = 2.98

Standard deviation (sd) = 0.05

A) proportion of cases with mean less than 3.014 inches

X = 3.014

Zscore = (x - m) / sd

Zscore = (3.014 - 2.98) / 0.05

= 0.034 / 0.05

= 0.68

P(Z < 0.68) ; from z table = 0.7517

B.)

25th percentile (0.25) = Q1; corresponds to a Zscore of about - 0.675

Zscore = (x - m) / sd

-0.675 = (X - 2.98) / 0.05

-0.675 * 0.05 = X - 2.98

-0.03375 = X - 2.98

X = 2.946 = 2.95

C)

Q3 = 0.75

75th percentile (0.75) ; corresponds to a Zscore of about 0.675

Zscore = (x - m) / sd

0.675 = (X - 2.98) / 0.05

0.675 * 0.05 = X - 2.98

0.03375 = X - 2.98

0.03375 + 2.98 = X

X = 3.01

IQR = Q3 - Q1

IQR = 3.01 - 2.95 = 0.06

Lower outlier :

Q1 - 1.5(IQR) = 2.95 - 1.5(0.06) = 2.86

UPPER OUTLIER:

Q3 + 1.5(IQR) = 3.01 + 1.5(0.06) = 3.10

Proportion less than 2.86:

Z = (2.86 - 2.98) / 0.05

= - 0.12 / 0.05

= - 2.4

P(Z < 2.4) = 0.0082

Proportion < 3.10

Z = (3.1 - 2.98) / 0.05

= 0.12 / 0.05

= 2.4

P(Z > 2. 4) = 1 - P(Z < 2.4) = (1 - 0.9918) = 0.0082

Proportion of

i agree because i see the other persons thinking is clear