1. ## Normal Distribution

I hope I'm posting this in the right place!

I've taken on a summer maths course to earn some extra credits for my course and although I've been staring at the information all day, I still have no idea how to answer any of the questions properly.

Would someone mind taking me through one of the questions?

The length of time, in minutes, needed to play 18 holes of golf at a municipal golf course on Saturday is approximately normally distributed, with a mean of 240 minutes and a standard deviation of 45 minutes. If you set out to play 18 holes of golf on this course on a Saturday, what percentage of the time will it take you the following lengths of time to finish?

(a) More than 5 hours
(b) Less than 4 hours
(c) Between 210 minutes and 320 minutes
(d) You tee off at 9am; would it be sensible to arrange to meet a
friend for lunch at 12noon ?

Thanks

2. Originally Posted by MissM

The length of time, in minutes, needed to play 18 holes of golf at a municipal golf course on Saturday is approximately normally distributed, with a mean of 240 minutes and a standard deviation of 45 minutes. If you set out to play 18 holes of golf on this course on a Saturday, what percentage of the time will it take you the following lengths of time to finish?

(a) More than 5 hours
(b) Less than 4 hours
(c) Between 210 minutes and 320 minutes
(d) You tee off at 9am; would it be sensible to arrange to meet a
friend for lunch at 12noon ?

Thanks

You need to convert these questions into z-scores.

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

$\displaystyle \mu = 240$
$\displaystyle \sigma = 45$

For part a)

$\displaystyle x = 300$ (5 hours)

$\displaystyle P(X>300) = P\left(Z>\frac{300-240}{45}\right) = P(Z>1.33)$

For part b)

$\displaystyle x = 240$ (5 hours)

$\displaystyle P(X<240) = P\left(Z<\frac{240-240}{45}\right) = P(Z<0)$

To finish the problems you will need a normal cdf table.

Does this make sense?