Question

A random sample of size 100 is taken from a population with the mean as 76 and the variance as 256. what is probability of getting sample mean:
a.between 74.4 and 78.4,
b.less than 74.4, and
c.greater than 78.4.

Answer 1

Given the population mean is $\mu=76$ and population standard deviation is $\sigma=\sqrt{256} =16$.

Now sample mean is $\mu_X=76$ and standard deviation of the sample mean is $\sigma_X=\frac{\sigma}{\sqrt{n}} =\frac{16}{\sqrt{100}}=1.6$.

Use Z-score for finding the probabilities. Use a standard normal distribution table.

a)The probability

$P(74.4<X<78.4)=P(\frac{74.4-76}{1.6}<Z<\frac{78.4-76}{1.6}) \\ P(74.4<X<78.4)=P(-1<Z<1.5) \\ P(74.4<X<78.4)=0.7745$

b)The probability

$P(X<74.4)=P(Z<\frac{74.4-76}{1.6}) \\ P(X<74.4)=P(Z<-1) \\

P(X<74.4)=0.1587

$

c)The probability

$P(X>78.4)=P(Z>\frac{78.4-76}{1.6}) \\ P(X>78.4)=P(Z>1.5) \\

P(X>78.4)=0.0668

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