Q. 3.142 - Elementary Statistics

Question

 PGA Driving Distances. The PGA TOUR provides various statistics on performance of players in the Prof Association of America. For the week ending september $9,2013$, the year-to-date leader for longest average drive was B. his $773$ drives, he averaged $298.5$ yards. Assume Standard deviation of $19.7$ yards.

a. Construct a graph
b. Apply Property $1$ of the empirical rule to make pertinent statements about the observations in the sample.
c. Repeat part (b) for Property $2$ of the empirical rule.
d. Repeat part (b) for Property $3$ of the empirical rule.

Step-by-Step Solution

Answer

Part(a) Required graph is given below.

Part(b) By property $1$ of empirical rule $68\%$ of sample to make statement about observation in sample.

Part(c)  By property $2$ of empirical rule $95\%$ of sample to make statement about observation in sample.

Part(d)  By property $3$ of empirical rule $99.7\%$ of sample to make statement about observation in sample.

Part(a) Step 1 : Given information

We are given that the leader for longest average drive was B. his $773$ drives, he averaged $298.5$  yards. with standard deviation of $19.7$ yards.

Part(a) Step 2 : Simplify

For drawing the graph, we need to find standard deviation on either side of mean.

It means , we need to find $\overline{x}-3s, \overline{x}-2s, \overline{x}-s, \overline{x}, \overline{x}+s, \overline{x}+2s, \overline{x}+3s$

Which is discussed in Step $4,6,8$

Required graph is

Part(b) Step 1 : Given information

We are given that the leader for longest average drive was B. his $773$ drives, he averaged $298.5$ yards. with standard deviation of $19.7$ yards.

Part(b) Step 2 : Simplify

As we already know by Property I of the empirical rule, nearly $68\%$ of the player B in the sample have leader for longest drive within one standard deviation to either side of the mean.

 Now,

$773\times \frac{68}{100}=525.64\approx 526$

One standard deviation to either side of the mean is from
$$\begin{gathered} \overline{x}-s=298.5-19.7=278.8 \\ \overline{x}+s=298.5+19.7=318.2 \end{gathered}$$

From above calculations, we can say that,

Approximately $526$ of the $773$ drives in the sample have the year-to-date leader between $278.8$ and $318.2$ yards.

Part(c) Step 1 : Given information

We are given that the leader for longest average drive was B. his $773$ drives, he averaged $298.5$ yards. with standard deviation of $19.7$ yards.

Part(c) Step 2 : Simplify

As we already know by Property 2 of the empirical rule, nearly $95\%$ of the men in the sample have forearm lengths within two standard deviations to either side of the mean.

Now, 

$773\times \frac{95}{100}=734.35\approx 734$

 Two standard deviations to either side of the mean is from
$$\begin{gathered} \overline{x}-2s=298.5-\left(2\right)19.7=259.1 \\ \overline{x}+2s=298.5+\left(2\right)19.7=337.9 \end{gathered}$$

From above calculations , we can interpret that,

Approximately $734$ of the $773$ drives in the sample have year-to-date leader between $259.1$ and $337.9$ yards

Part(d) Step 1 : Given information

We are given that the leader for longest average drive was B. his $773$ drives, he averaged  $298.5$ yards. with standard deviation of $19.7$ yards.

Part(d) Step 2 : Simplify

As we know by Property 3 of the empirical rule, approximately $99.7\%$of the drives in the sample have leader within three standard deviations to either side of the mean.

Now, 

$773\times \frac{99.7}{100}=770.68\approx 771$

Three standard deviations to either side of the mean is from
$$\begin{gathered} \overline{x}-3s=298.5-\left(3\right)19.7=239.4 \\ \overline{x}+3s=298.5+\left(3\right)19.7=357.6 \end{gathered}$$

From above calculations, we can say that

Approximately $771$ of the $773$ drives in the sample have year-to-date leader  between $239.4$ and $357.6$ yards