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The Random Variable is normally distributed with mean = 98 and standard deviation =
20.41.
On the sampling distribution with n = 81, for what value of the sample mean (X) would have
10% of the possible sample means less than it? Round your answer to 2 decimal places.

Respuesta :

If mean is 98, standard deviation is 20.41, n=81 then sample mean will be  131.57445 which would have 10% of the possible mean less than it.

Given that mean is 98,standard deviation is 20.41,n=81 and significance level be 10%.

We are required to find the value of sample mean.

Because the value of n is greater than 30 so we will use z statistics in the problem.

We know that,

Z=(X-μ)/σ  in which μ is population mean, σ is standard deviation.

Confidence level=1-0.01=0.09 means 90%.

Z value for 90% confidence level be 1.645.

we have to put the value of Z=1.645 , μ=98,σ=20.41 to get the value of X.

1.645=(X-98)/20.41

1.645*20.41=X-98

33.57445=X-98

X=98+33.57445

X=131.57445

Hence if mean is 98, standard deviation is 20.41, n=81 then sample mean will be  131.57445 which would have 10% of the possible mean less than it.

Learn more about z test at https://brainly.com/question/14453510

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