Question

Note: For any part below, if it cannot be answered, enter NA.

A quality control company was hired to study the length of meter sticks produced by a certain company. The team carefully measured the length of many many meter sticks, and the distribution seems to be slightly skewed to the right with a mean of 100.03 cm and a standard deviation of 0.13 cm.

a) What is the probability of finding a meter stick with a length of more than 100.12 cm?

b) What is the probability of finding a group of 22 meter sticks with a mean length of less than 100 cm?

c) What is the probability of finding a group of 34 meter sticks with a mean length of more than 100.06 cm?

d) What is the probability of finding a group of 44 meter sticks with a mean length of between 100.02 and 100.05 cm?

e) For a random sample of 50 meter sticks, what mean length would be at the 92nd percentile?

Answer 1

This is a normal distribution question with

a) P(x > 100.12)=?
The z-score at x = 100.12 is,

z = 0.6923
This implies that
P(x > 100.12) = P(z > 0.6923) = 1 - 0.7556


b) Sample size (n) = 22
Since we know that


P(x < 100.0)=?
The z-score at x = 100.0 is,

z = -1.083
This implies that


c) Sample size (n) = 34
Since we know that

P(x > 100.06)=?
The z-score at x = 100.06 is,

z = 1.3453
This implies that
P(x > 100.06) = P(z > 1.3453) = 1 - 0.9107358132109504


d) Sample size (n) = 44
Since we know that

P(100.02 < x < 100.05)=?

This implies that
P(100.02 < x < 100.05) = P(-0.5102 < z < 1.0204) = P(Z < 1.0204) - P(Z < -0.5102)
P(100.02 < x < 100.05) = 0.8462306030582083 - 0.30495567609512414


e) Sample size (n) = 50
Since we know that

Given in the question
P(X < x) = 0.92
This implies that
P(Z < 1.4051) = 0.92
With the help of formula for z, we can say that

x = 100.0559
PS: you have to refer z score table to find the final probabilities.
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