Question

The standard deviation of a population is 2.70 cms. Find the probability that is
random sample of size 66 (i) the sample mean will differ from the population mean
by 0.75 cm or more, and (ii) the sample mean will exceed the population mean by
0.75 cm. or more (given that the value of the standard normal probability integral
from 0 to 2.25 is 0.4877).

Answer 1

Let $\overline{x}$ be the sample mean and $\mu$ be the population mean. We have $\sigma=2.7$ (standard deviation of the population), $n=66$ (size of the sample). We should find $P(|\overline{x}-\mu|\geq 0.75)$ and $P(\overline{x}-\mu \geq 0.75)$.

Let $X_1,\ldots,X_{66}$ is our sample (each $X_i$ is a random variable). Consider $\overline{x}=\frac{1}{66}(X_1+\ldots+X_{66})$ (the sample mean $\overline{x}$ is a random variable).

Random variable $\overline{x}$ is normally distributed with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}=\frac{2.7}{\sqrt{66}}$.

Then random variable $\overline{x}-\mu$ is normally distributed with mean 0 and standard deviation $\frac{2.7}{\sqrt{66}}.$

ii) $P(\overline{x}-\mu\geq 0.75)=P(Z\geq\frac{0.75}{\frac{2.7}{\sqrt{66}}})=1-P(Z\leq 2.25)=1-0.4877=0.5123.$

i)

i$P(|\overline{x}-\mu|\geq 0.75)=P(|Z|\geq\frac{0.75}{\frac{2.7}{\sqrt{66}}})=1-P(|Z|\leq\frac{0.75}{\frac{2.7}{\sqrt{66}}})=\\ =1-P(-\frac{0.75}{\frac{2.7}{\sqrt{66}}}\leq Z\leq\frac{0.75}{\frac{2.7}{\sqrt{66}}})=1-(0.4877+0.4877)=0.0246.$

Random variable Z has a standard distribution.