At the end of the spring semester, the Dean of Students sent a survey to the entire freshman class. One question asked the students how much weight they had gained or lost since the beginning of the school year. The average was a gain of $\mu = 9$ pounds with a standard deviation of $\sigma = 6$. The distribution of scores was approximately normal. A sample of $n = 4$ students is selected and the average weight change is computed for the sample.

a. What is the probability that the sample mean will be greater than M = 10 pounds? In symbols, what is $P(M > 10)$?

b. Of all of the possible samples, what proportion will show an average weight loss? In symbols, what is $P(M < 0)$?

c. What is the probability that the sample mean will be a gain of between $M = 9$ and $M = 12$ pounds? In symbols, what is $P(9 < M < 12)$?

#### Solution

Given that $\mu = 9$ and $\sigma = 6$.

a. The probability that the sample mean will be greater than M = 10 pounds is

` $$ \begin{aligned} P(M > 10) &= P\big(\frac{M-\mu}{\sigma/\sqrt{n}} > \frac{10-9}{6/\sqrt{4}}\big)\\ & =P(Z > \frac{0-9}{3}\big)\\ & =P(Z > -3)\\ & =1- P(Z < -3)\\ &= 1-0.067\\ &= 0.933 \end{aligned} $$ `

b. The proportion will show an average weight loss is

` $$ \begin{aligned} P(M < 0) &= P\big(\frac{M-\mu}{\sigma/\sqrt{n}} < \frac{0-9}{6/\sqrt{4}}\big)\\ & =P(Z < \frac{0-9}{3}\big)\\ & =P(Z < -3)\\ &= 0.067 \end{aligned} $$ `

c. The probability that the sample mean will be a gain of between $M=9$ and $M=12$ pounds

` $$ \begin{aligned} P(9 < M < 12) &= P(9 < M < 12)\\ &= P\big(\frac{9-9}{6/\sqrt{n}} < \frac{M-\mu}{\sigma/\sqrt{n}} < \frac{12-9}{6/\sqrt{ n}}\big)\\ &=P(0 < M < 1\big)\\ &=P(Z < 1)- P(Z < 0)\\ &=0.691 - 0.5\\ &=0.191\\ \end{aligned} $$ `

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators