b) 38.3% c)0.0049cm PROBLEM 3: [10 points] Consider a company that produces hard

Question

b) 38.3%

c)0.0049cm

PROBLEM 3: [10 points] Consider a company that produces hard shell covers for smartphones. To achieve a preferred fit between the cover and the smartphone, the length of the cover needs to be between 8.99cm and 9.01cm. What is the likelihood that the length of the cover will be between these limits, a) If the length of the cover is a random variable that is uniformly distributed between 8.95cm and 904cm? b) If the length of the cover is a random variable that follows a normal distribution with a mean of 9.00cm and a standard deviation of 0.02cm? c) Assume the length of the cover is a random variable that follows a normal distribution with a mean of 9.00cm. Then what is the required standard deviation of the production process to ensure that the length of the 2% of all covers is larger than 9.01 cm?

Explanation / Answer

a)here for uniform distribution ; paramter a=8.95 and b=9.04

hence CDF =P(X<x) =(x-a)/(b-a) =(x-8.95)/(9.04-8.95)=(x-8.95)/0.09

hence P(length between 8.99 and 9.01)=P(8.99<X<9.01)=P(X<9.01)-P(X<8.99)

=(9.01-8.95)/0.09-(8.99-8.95)/0.09=0.02/0.09 =2/9 =0.2222 ~ 22.2%

b)

c)

here for top 2 percentile ; critical z =2.05

hence std deviaiton =(X-mean)/z score =(9.01-9)/2.05 =0.0049

for normal distribution z score =(X-)/x here mean=       = 9 std deviation   == 0.020

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