Question

find the probability that the mean of a sample of 25 people is less than 10. p(\overline{x}<10)=0.0016 part 2 of 3 find the probability that the mean of a sample of 25 people is greater than 10. p(\overline{x}>10)=0.9984 part 3 of 3 find the probability that the mean of a sample of 25 people is between 11 and 12. p(11<\overline{x}<12)=

Explanation:

Step1: Calculate the z - scores

Assume the population mean is \(\mu\) and the population standard deviation is \(\sigma\). For a sample of size \(n = 25\), the standard error of the mean is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\).
The z - score formula is \(z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}\).
For \(\bar{x}_1 = 11\), \(z_1=\frac{11 - \mu}{\sigma_{\bar{x}}}\).
For \(\bar{x}_2 = 12\), \(z_2=\frac{12 - \mu}{\sigma_{\bar{x}}}\).

Step2: Use the standard normal distribution table

\(P(11<\bar{X}<12)=P(z_1<Z<z_2)\)
\(=P(Z < z_2)-P(Z < z_1)\)

Answer:

Assuming \(\mu = 10\) and \(\sigma = 5\) (since they are not given in the problem, these are common values for illustration purposes. If actual values of \(\mu\) and \(\sigma\) are provided, substitute them in the following steps).
\(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{5}{\sqrt{25}} = 1\)
For \(\bar{x}=11\), \(z_1=\frac{11 - 10}{1}=1\)
For \(\bar{x}=12\), \(z_2=\frac{12 - 10}{1}=2\)
\(P(11<\bar{X}<12)=P(Z < 2)-P(Z < 1)\)
From the standard normal distribution table, \(P(Z < 2)=0.9772\), \(P(Z < 1)=0.8413\)
\(P(11<\bar{X}<12)=0.9772 - 0.8413=0.1359\)