Answer to Question #243431 in Statistics and Probability for dinesh

Question

An apple juice producer buys apples from a large apple grove that has one variety of apple. The amount of juice squeezed from these apples is approximately normally distributed, with a mean of 4.70 ounces and a standard deviation of 0.40 ounce. Suppose that you select a sample of 25 apples

a) The probability is 70% that the sample mean amount of juice will be contained between what two values symmetrically distributed around the population mean?

b) The probability is 77% that the sample mean amount of juice will be greater than what value?

Answer 1

Let $X=$ the sample mean amount of juice: $X\sim N(\mu, \sigma^2/n).$

Given $\mu=4.70,\sigma=0.40,n=25.$

a)

$$P(\mu-a<X<\mu+a)=1-2P(X<\mu-a)$$

$$=1-2P(Z<\dfrac{\mu-a-\mu}{\sigma/\sqrt{n}})=1-2P(Z<\dfrac{-a}{\sigma/\sqrt{n}})$$

$$=0.70$$

$$P(Z<\dfrac{-a}{\sigma/\sqrt{n}})=0.15$$

$$\dfrac{-a}{\sigma/\sqrt{n}}=-1.036433$$

$$a=1.036433(\dfrac{0.40}{\sqrt{25}})=0.08291464$$

$$\mu-a=4.70-0.083=4.617$$

$$\mu+a=4.70+0.083=4.783$$

$$P(4.617<X<4.783)=0.70$$

b)

$$P(X>x_2)=1-P(X\leq x_2)$$

$$=1-P(Z<\dfrac{x_2-\mu}{\sigma/\sqrt{n}})=0.77$$

$$P(Z<\dfrac{x_2-\mu}{\sigma/\sqrt{n}})=0.23$$

$$\dfrac{x_2-\mu}{\sigma/\sqrt{n}}=-0.738847$$

$$x_2=4.70-0.738847(\dfrac{0.40}{\sqrt{25}})=4.641$$

$$P(X>4.641)=0.77$$