Question

Let the random variable X follow a Normal distribution with variance σ2 = 625. A random...

Let the random variable X follow a Normal distribution with variance σ2 = 625.

A random sample of n = 50 is obtained with a sample mean, X-Bar of 180.

What is the probability that μ is between 198 and 211?

What is Z-Score1 for μ greater than 198?

Homework Answers

Answer #1

Table in MS-Excel
sigma2= 625
sigma= 25
sqrt(2)= 1.414214
n 50
sqrt(n)= 7.071068
xbar 180
xbar-198 -18
xbar-211 -31
(xbar-198)/(sigma/sqrt(n)) -5.09117
(xbar-211)/(sigma/sqrt(n)) -8.76812
P(Z<-8.768124) 9.08E-19
P(Z<-5.091169) 1.78E-07
ROUND(NORMSDIST(-8.768124),1) 0
ROUND(NORMSDIST(-5.091169),1) 0

In general,a z-score indicates how many standard deviations an element is from the mean. It can be calculated from the following formula: z = (X - μ) / σ

where, z is the z-score, X is the value of the element, μ is the population mean, and σ is the standard deviation.

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