An anthropologist wishes to estimate the average height of men for a certain race of people. If the population standard deviation is assumed to be 2.8 inches and if she randomly samples 100 men, find the probability that the difference between the sample mean and the true population mean will not exceed 0.7 inch. (Round your answer to four decimal places.)

The probability that the difference between the sample mean [tex]\bar{x}[/tex] and the true population mean [tex]\mu[/tex] is : [tex]P[|\bar{x}-\mu|\leq0.7]=0.9876[/tex]

Step-by-step explanation:

Given :

Population standard deviation [tex]\sigma=2.8[/tex].

Sample size [tex]n=100[/tex]

The sample mean standard deviation[tex]=\frac{\sigma}{\sqrt{n}}[/tex]

To find :

The probability that the difference between the sample mean [tex]\bar{x}[/tex] and the true population mean [tex]\mu[/tex]

[tex]\because\frac{\sigma}{\sqrt{n}}=\frac{2.8}{\sqrt{100}}=\frac{2.8}{10}=0.28[/tex]

Now, the probability that the difference between the sample mean [tex]\bar{x}[/tex] and the true population mean [tex]\mu[/tex] is :

[tex]P[|\bar{x}-\mu|\leq0.7][/tex], [tex]n[/tex]is large [tex]\Rightarrow \frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}=z[/tex]

[tex]\Rightarrow P[\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{0.7}{\frac{\sigma}{\sqrt{n}}}]=P[z<\frac{0.7}{0.28}]=P[z<2.5][/tex]

[tex]\Rightarrow P[z<2.5]=P[-2.5<z<2.5][/tex]

[tex]\Rightarrow P[-2.5<z<2.5]=P[z<2.5]-P[z\leq-2.5]\\\Rightarrow P[z<2.5]-P[z\leq-2.5]=2P[z\leq2.5]-1\\\Rightarrow 2P[z\leq2.5]-1=0.9876[/tex]

Therefore, the probability [tex]P[|\bar{x}-\mu|\leq0.7]=0.9876[/tex]