# Normal distribution question

• Oct 29th 2008, 01:34 PM
Panzz
Normal distribution question
I can't seem to be able to understand word problems

an applicant take a test . The time it normally distributed with mean time 60 minutes and standard deviation 8 minutes . Determine how much time an applicant should be given in general so 97 percent of them will be able to complete the test

(Bow)(Bow)(Bow)
• Oct 31st 2008, 10:42 AM
BoboStrategy
Re: Normal dist
First step back and think about it. The mean time is 60, and you want to find the time that 97% of folks take less than. "97%" corresponds to a little more than 2 standard deviations above the mean, right? [Why?]
So your answer will be a little less than 60 + 2 * 8 = 76...right?

Here's a hint on the details:

You want to find time T so that F(t) = prob(t < T) = 0.97.

We know
$
F(t) = \phi((t - \mu)/ \sigma)
$

So you want to solve
$
0.97 = \phi((t - 60)/ 8)
$

for t.

We know
$
\phi(1.88) = 0.97
$

So you want to find t so that
$
(t - 60)/ 8 = 1.88
$
• Nov 1st 2008, 12:39 AM
Angel
Angel (BestSolution)
Let “x” variable is normally distributed, represents the time.

µ=60 and σ=8 are the mean and standard deviation of the time.

So we have to find x1 where 97% students will be able to complete their test.

So,

z1 = (x1 - µ)/σ (z1 lies on z-axis corresponding x1 on x-axis)

x1=µ+z1σ
x1= 60+(1.88)(8) [z1 take from z-dist. Table]
x1= 75.04 ~75
x1= 75

So if the time will be 75 min, then the students will be able to complete their test 97% complete.