# Stat question

• December 30th 2008, 11:53 PM
Stat question
Golf balls must meet a set of five standards in order to be used in professional tournaments. One of these standards is the distance traveled. When a ball is hit by a mechanical device, Iron Byron, with a 10-degree angle of launch, a backspin of 42 revolutions per second, and a ball velocity of 235 feet per second, the distance the ball travels may bot exceed 291.2 yards. Manufacturers want to develop balls that will travel so close to the 291.2 yards as possible without exceeding that distance. A particular manufacturer has determined that the distances traveled for the balls it produces are normally distributed with a standard deviation of 2.8 yards. This manufacturer has a new process that allows it to set the mean distance the ball will travel.

1.) If the manufacturer sets the mean distance traveled to be equal to 288 yards, what is the probability that a ball that is randomly selected for testing will travel too far?

2.)Assume the mean distance traveled is 288 yards and that five balls are independently tested. What is the probability that at least one of the five balls will exceed the maximum distance of 291.2 yards?

3.) If the manufacturer wants to be 99 percent certain that a randomly selected ball will not exceed the maximum distance of 291.2 yards, what is the largest mean that can be used in the manufacturing process?

No ideas? wanna show me whos boss lol :P thanks in advanced!
• December 31st 2008, 12:26 AM
nzmathman
If you are wanting to use tables to solve the problems,
remember for the standard normally distributed variable $Z$ with $\sigma = 1$ and $\mu = 0$, $Z = \frac{X - \mu}{\sigma}$.

Let X be the distance the golf ball travels.

So, (1):

$P(X > 291.2) = P\left(Z > \frac{291.2 - 288}{2.8}\right) = P\left(Z > \frac{3.2}{2.8}\right) = P(Z > 1.14)$

Now you can find this probability - the event the ball goes further than allowed.

For number (2), this requires the binomial distribution. Let Y be the event a ball goes further than allowed. The number of trials, n, is 5, and we want $P(Y \geq 1) = 1 - P(Y = 0)$, where

$P(Y = 0) = \binom{5}{0} \cdot [P(Y)]^0 \cdot [1 - P(Y)]^5$

(3) We want $P(X \leq 291.2) = 0.99$

Now as in Q1, convert this to a probability for $Z$. Now looking at a normal distribution table, we can see $P(Z < 2.326) = 0.99$.

Now you will have an equation of the form $0.99 = \frac{X - \mu}{\sigma}$where you know $X$ and $\sigma$ so you can find $\mu$

Can you take it from here?
• December 31st 2008, 04:56 PM
Im confused on number 1. I looked in my book and cant figure it out. My teacher expects us to go over this stuff over break, and have a test first day back. I really am lost.
• December 31st 2008, 06:02 PM
nzmathman
One of the first things you learn about normal distribution is the standard variable Z, and as I have stated in my first post, how to find a Z score corresponding to another variable.

You want $P(X > 291.2)$, which can be converted to $P(Z > 1.143)$.

To find this probability using tables:
The table gives values for $P(0 < Z < k)$. Looking up 1.143 for k, we see that $P(0 < Z < 1.143) = 0.3735$.

But we want $P(Z > 1.143)$. We know that $P(Z < 0) =0.5$ and that $P(0 < Z < 1.143) = 0.3735$.

So logically, $P(Z > 1.143) = 1 - 0.5 - 0.3735 = 0.1265.$
• December 31st 2008, 07:34 PM
for 1 --- Oh so, the probabilty that a randomly selected ball will travel to far or ( >1.143 ) is 12.65% ?

for 2 ---- I understand $P(Y \geq 1) = 1 - P(Y = 0)$

But get lost here :
$P(Y = 0) = \binom{5}{0} \cdot [P(Y)]^0 \cdot [1 - P(Y)]^5$

for 3 --- i dont get how you got
$P(Z < 2.326) = 0.99$
Because you still do the same as in Q1 but you got a different number.

Im sorry if you feel like im wasting your time but im learning a lot here and DO appreciate the help.

• December 31st 2008, 08:33 PM
nzmathman
For number 2, we have 5 balls and we want to find the probability that 0 go past 291.2 yards. This situation is a binomial distribution - fixed number of trials, fixed probability of success, and there are only 2 possible outcomes. I have simply substituted n=5 and k=0 into the binomial distribution formula here.

I'll get back to you on #3 soon, as I have to go for a while.
• December 31st 2008, 08:48 PM
Thanks for the help, i see where you got 5 from now, but im confused on how to get an answer from that equation. Ill look up binomial distribution for af ixed number of trials
• December 31st 2008, 08:50 PM
nzmathman
Ok, number 3 - the manufacturer wants to be 99% certain the balls do not go too far. Because we are trying to find a new value for the mean, we must start with the Z score first and convert to a probability in X. Either by learning confidence intervals or by looking at a normal distribution table, you can deduce that $P(Z < 2.326) = 0.99$.

Remember $Z = \frac{X - \mu}{\sigma}$

We know that $X = 291.2$, $\sigma = 2.8$, and we are wanting to find $\mu$

$\therefore \frac{X - \mu}{\sigma} = 2.326$

$\Rightarrow \frac{291.2 - \mu}{2.8} = 2.326$

$\Rightarrow 291.2 - \mu = 2.326 \times 2.8$

Now it's just a matter of you finding $\mu$, the new mean.
• December 31st 2008, 09:35 PM
nzmathman
Quote:

${5\choose0} = {}^5 C_0$ (combinations)
How many ways to pick 0 objects from 5 objects? Only one. So ${5\choose0} = 1$.
Any number to power of 0 is equal to 1. So $[P(Y)]^0 = 1$
Now all this leaves is $[1 - P(Y)]^5$
Remember we calculated $P(Y)$ in Q1 $= 0.1265$.
So $[1 - P(Y)]^5 = (1 - 0.1265)^5$