USGA golf ball specifications. According to the US. Golf Association (USGA), "The diameter of the [golf] ball must not be less than 1.680 inches" (USGA, 2016). The USGA periodically checks the specifications of golf balls by randomly sampling balls from pro shops around the country. Two dozen of each kind are sampled, and if more than three do not meet requirements, that kind of ball is removed from the USGA's approved-ball list.

1. What assumptions must be made and what information must be known in order to use the binomial probability distribution to calculate the probability that the USGA will remove a particular kind of golf ball from its approved-ball list?
2. Suppose 10% of all balls produced by a particular manufacturer are less than 1.680 inches in diameter and assume that the number of such balls, x, in a sample of two dozen balls can be adequately characterized by a binomial probability distribution. Find the mean and standard deviation of the binomial distribution.
3. Refer to part b. If x has a binomial distribution, then so does the number, y, of balls in the sample that meet the USGA's minimum diameter. [Note: x + y = 24] Describe the distribution of y. In particular, what are p, q and n? Also, find E(y) and the standard deviation of y.

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" $$\begin{gathered} \mathrm{n}=2\times 12 \\ =24 \end{gathered}$$

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" $$\begin{gathered} \mathrm{Mean},\mathrm{\mu }=\mathrm{np} \\ =24\times 0.10 \\ =2.4 \end{gathered}$$

The standard deviation of x is computed as:

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

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

The standard deviation of y is computed as:

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

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.