Question 1204553: Suppose that battery lives are normally distributed with a mean of 12.89 hours and a standard deviation of 1.47 hours. What is the minimum sample size that would be required so that the probability of obtaining a sample mean above 13.25 hours is less than 1%?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
mu = population mean = 12.89
sigma = population standard deviation = 1.47
xbar = sample mean = 13.25
n = sample size = unknown
Let's consider the situation where
P(xbar > 13.25) = 0.01
That would be equivalent to
P(Z > c) = 0.01
where the z score is calculated like so
z = (xbar - mu)/( sigma/sqrt(n) )
P(Z > c) = 0.01 solves to c = 2.326 approximately through use of a stats calculator such as this one
https://davidmlane.com/hyperstat/z_table.html
or use the invNorm function on a TI83 calculator.
A table is another option if your teacher won't let you use a calculator. The drawback with a table is that it's far less accurate.
Use this z score to determine the sample size
z = (xbar - mu)/( sigma/sqrt(n) )
2.326 = (13.25 - 12.89)/( 1.47/sqrt(n) )
2.326 = 0.36/( 1.47/sqrt(n) )
2.326*( 1.47/sqrt(n) ) = 0.36
2.326*(1.47) = 0.36*sqrt(n)
sqrt(n) = (2.326*1.47)/0.36
n = (2.326*1.47/0.36)^2
n = 90.2088
n = 91
We round up to the nearest integer even though 90.2088 is closer to 90 than it is to 91.
We round up to clear the threshold needed.
If n = 90, then,
z = (xbar - mu)/( sigma/sqrt(n) )
z = (13.25 - 12.89)/( 1.47/sqrt(90) )
z = 2.32330603604208
z = 2.323
Use a Z calculator to find that P(Z > 2.323) = 0.01009 = 1.009% approximately, which does not clear the threshold of "less than 1%".
If n = 91, then,
z = (xbar - mu)/( sigma/sqrt(n) )
z = (13.25 - 12.89)/( 1.47/sqrt(91) )
z = 2.33617763612312
z = 2.336
Use a Z calculator to find that P(Z > 2.336) = 0.00975 = 0.975% approximately, which does clear the threshold of "less than 1%".
Answer: 91
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