Chapter 7.3, Problem 75E

To determine

To find: that the sample mean will fall in which of the following intervals.

Answer to Problem 75E

15.95 to 16.15 ounces

Explanation of Solution

Given:

  $\begin{array}{c}\mu =16.05\\ \sigma =0.1\\ n=4\end{array}$

Formula used:

  $z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}$

Calculation:

Since the population distribution is normal, the sampling distribution of the sample mean $\overline{x}$ is also normal.

The Z-score is

  $z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{15.85-16.05}{0.1/\sqrt{4}}=-4.00$

  $z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{15.90-16.05}{0.1/\sqrt{4}}=-3.00$

  $z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{15.95-16.05}{0.1/\sqrt{4}}=-2.00$

  $z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{16.00-16.05}{0.1/\sqrt{4}}=-1.00$

  $z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{16.05-16.05}{0.1/\sqrt{4}}=0.00$

  $z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{16.10-16.05}{0.1/\sqrt{4}}=1.00$

  $z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{16.15-16.05}{0.1/\sqrt{4}}=2.00$

  $z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{16.20-16.05}{0.1/\sqrt{4}}=3.00$

  $z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{16.25-16.05}{0.1/\sqrt{4}}=4.00$

The associating probability using the normal probability table

  $P\left(Z<-3.00\right)$ is given in the row starting with $-3.0$ and in the column starting with $.00$ of the standard normal probability table in the appendix. Similar for the other probabilities

a)

  $\begin{array}{c}P\left(16.05<\overline{X}\ge 16.15\right)=P\left(0.00<Z<2.00\right)\\ =P\left(Z<2.00\right)-P\left(Z<0.00\right)\\ =0.9772-0.05000\\ =0.4772\\ =47.72\%\end{array}$

b)

  $\begin{array}{c}P\left(16.00<\overline{X}\ge 16.10\right)=P\left(-1.00<Z<1.00\right)\\ =P\left(Z<1.00\right)-P\left(Z<-1.00\right)\\ =0.8413-0.1587\\ =0.6826\\ =68.26\%\end{array}$

c)

  $\begin{array}{c}P\left(15.95<\overline{X}\ge 16.15\right)=P\left(-2.00<Z<2.00\right)\\ =P\left(Z<2.00\right)-P\left(Z<-2.00\right)\\ =0.9772-0.0228\\ =0.9544\\ =95.44\%\end{array}$

d)

  $\begin{array}{c}P\left(15.95<\overline{X}\ge 16.15\right)=P\left(-2.00<Z<2.00\right)\\ =P\left(Z<2.00\right)-P\left(Z<-2.00\right)\\ =0.9772-0.0228\\ =0.9544\\ =95.44\%\end{array}$

e)

  $\begin{array}{c}P\left(15.90<\overline{X}\ge 16.20\right)=P\left(-3.00<Z<3.00\right)\\ =P\left(Z<3.00\right)-P\left(Z<-3.00\right)\\ =1-0\\ =1\\ =100\%\end{array}$

We note that the probability is closest to 95% is $P\left(15.95<\overline{X}<16.15\right)$

Hence, the correct option is (c)