91Ó°ÊÓ

The golf ball must have a diameter of at least 1.680 inches.

The sampled golf balls are two dozen.

The assumptions are given below:

• The probability of completion stays constant from trial to trial.

• The practice is also independent of each other.

To use the binomial probability distribution, it is essential to know whether the ball meets the specific requirements or not.

Assume the number of golf balls is x.

i.e., $\mathrm{x}~\mathrm{B}(\mathrm{n},\mathrm{p})$

A binomial probability distribution can appropriately define a sample of two dozen balls. This data suggests that

The number of samples is obtained as:

localid="1664196263382" $$\begin{gathered} \mathrm{n}=2×12 \\ =24 \end{gathered}$$

10% of all balls manufactured by a specific company have a diameter of fewer than 1.680 inches.

The probability of success is obtained as:

localid="1664196275477" $$\begin{gathered} \mathrm{p}=\frac{10}{100} \\ =0.10 \end{gathered}$$

Therefore,

The mean of x is computed as:

localid="1664196244194" $$\begin{gathered} \mathrm{Mean},\mathrm{μ}=\mathrm{np} \\ =24×0.10 \\ =2.4 \end{gathered}$$

The standard deviation of x is computed as:

localid="1664196288439" $$\begin{gathered} \mathrm{σ}=\sqrt{\mathrm{np}\left(1-\mathrm{p}\right)} \\ =\sqrt{24×0.10(1-0.10)} \\ =\sqrt{2.16} \\ =1.4697 \end{gathered}$$

Hence, the mean of x is 2.4, and the standard deviation of x is 1.4697.

The x has a Binomial distribution.

The y represents the ball in the sample that meets the USGA's minimum diameter requirements.

The number of samples is n=24. [Note: x+y=24]

From part b, 10% of all balls produced by a particular manufacturer are less than 1.680 inches in diameter. So, 90% of the ball in the sample meets the USGA's minimum diameter requirements.

Here, the probability of success is calculated as:

$$\begin{gathered} \mathrm{p}=\frac{90}{100} \\ =0.90 \end{gathered}$$

The probability of failure is calculated as:

$$\begin{gathered} q=1-p \\ =1-0.90 \\ =0.10 \end{gathered}$$

Therefore,

The distribution of y is, $y~\left(24,0.90\right)$

The value of mean (y) is calculated as:

localid="1664195967863" $$\begin{gathered} \mathrm{Mean},\mathrm{E}\left(\mathrm{y}\right)=\mathrm{np} \\ =24×0.90 \\ =21.6 \end{gathered}$$

The standard deviation of y is computed as:

$$\begin{gathered} \mathrm{σ}=\sqrt{\mathrm{npq}} \\ =\sqrt{24×0.90×0.10} \\ =\sqrt{2.16} \\ =1.4697 \end{gathered}$$

Hence, the distribution of y is $y~B(24,0.90)$, the mean of y is 21.6, and the standard deviation of y is 1.4697.