Question

1. Scores are normally distributed with a mean of 86 and a standard deviation of 14. What is the probability that a random student scored below 72?

2. Suppose a student measuring the boiling temperature of a certain liquid observes the readings (in degrees Celsius) 102.5, 101.7, 103.1, 100.9, 100.5, and 102.2 on 6 different samples of the liquid. He calculates the sample mean to be 101.82. If he knows that the standard deviation for this procedure is 1.3 degrees, what is the confidence interval for the population mean at a 95% confidence level?

Answer 1

1. Let $X$ denote the score of an exam: $X\sim N(\mu,\sigma^2)$

Then

$$Z={X-\mu \over\sigma}\sim N(0,1)$$

Given that $\mu=86, \sigma=14.$

$$P(X<72)=P(Z<{72-86 \over 14})=P(Z<-1)\approx$$ $$\approx0.1587$$

The probability that a random student scored below $72$ is $0.1587\ (15.87\%).$

2. A confidence interval for a population mean with a known population standard deviation is based on the conclusion of the Central Limit Theorem that the sampling distribution of the sample means follow an approximately normal distribution.

Hence the sample mean will have the distribution $N(\mu,\sigma^2/n).$

The confidence interval we should use, then, is

$$CI=\bar{x}\pm z_{\alpha/2}{\sigma \over \sqrt{n}}$$

Given that $\bar{x}=101.82,n=6,\sigma=1.3,\alpha=0.05$

$$z_{\alpha/2}=z_{0.05/2}=1.96$$

$$CI=101.82\pm 1.96\cdot{1.3 \over \sqrt{6}}$$

$$CI=(100.78,102.86)$$