Answer to Question #270285 in Statistics and Probability for Sonia

Question

A random sample of 16 observations is to be drawn from a normal distribution having mean 11 and standard deviation 3. Let  denote the sample mean. Find, correct to three decimal places

(i)                The probability that  will have a value between 9.2 and 12.2

(ii)              The value of c for which

Answer 1

Let $x$ denote the sample mean: $x\sim N(\mu, \sigma/\sqrt{n}).$

Given $\mu=11, \sigma=3, n=16$

(i)

$$P(9.2<x<12.2)=P(x<12.2)-P(x\leq 9.2)$$

$$=P(z<\dfrac{12.2-11}{3/\sqrt{16}})-P(z\leq\dfrac{9.2-11}{3/\sqrt{16}})$$

$$=P(z<1.6)-P(z\leq -2.4)$$

$$\approx0.0.9379452007-0.0081975=0.937003$$

(ii)

$$P(x<c)=0.03$$

$$P(z<\dfrac{x-11}{3/\sqrt{16}})=0.03$$

$$c=\dfrac{x-11}{0.75}\approx-1.880794$$

$$x\approx11-0.75(1.880794)$$

$$x\approx9.59$$