# standard normal2

• November 4th 2007, 02:31 AM
0123
standard normal2
An economics test is taken by a large group of students. The test scores are normally sidtrbuted with mean 70, and the probability that a randomly chones studnet receives a score less than 85 is 0. 9332.
Four students are chosen at random. What is the probability that at least one of them scores more than 80 points on this test.

How do you do with 4 people? Thanks so much.
• November 4th 2007, 03:10 AM
TheBrain
You will need to set up the normal distribution first and work out the chances that just 1 person scores more than 80 points.

Once you have gotten this, the probability that at least 1 out of 4 gets above 80, is just 1 - P(they all get less than 80).

If this doesn't help I will explain more.
• November 4th 2007, 04:30 AM
0123
Quote:

Originally Posted by TheBrain
- P(they all get less than 80).
.

How do I get it? Thank you
• November 4th 2007, 10:52 AM
CaptainBlack
Quote:

Originally Posted by 0123
An economics test is taken by a large group of students. The test scores are normally sidtrbuted with mean 70, and the probability that a randomly chones studnet receives a score less than 85 is 0. 9332.

This bit allows you to determin the standard deviation. 0.9332 is the
probability of observing a z of less than 1.5 from a standard normally
distributed random variable. So the z score corresponding to a mark
of 85 is:

$z = \frac{85-70}{\sigma} = 1.5$

so:

$\sigma=10$

Quote:

Four students are chosen at random. What is the probability that at least one of them scores more than 80 points on this test.

How do you do with 4 people? Thanks so much.
The probability p that at least 1 score more than 80 is 1 minus the probability
that all score less than 80. So if Q(80) is the probability that one score less
than 80, then:

$
p = 1-[Q(80)]^4
$

RonL
• November 4th 2007, 11:32 AM
0123
Quote:

Originally Posted by CaptainBlack
This bit allows you to determin the standard deviation. 0.9332 is the
probability of observing a z of less than 1.5 from a standard normally
distributed random variable. So the z score corresponding to a mark
of 85 is:

$z = \frac{85-70}{\sigma} = 1.5$

so:

$\sigma=10$

The probability p that at least 1 score more than 80 is 1 minus the probability
that all score less than 80. So if Q(80) is the probability that one score less
than 80, then:

$
p = 1-[Q(80)]^4
$

RonL

Uhm. Funny. Because there are quite a lot exercices like this and I did them differently getting the right result.
I used binomial distribution with P= (X> 80). Is it wrong? How is that we both come up with the right result?
Thanks a lot
• November 4th 2007, 12:09 PM
CaptainBlack
Quote:

Originally Posted by 0123
Uhm. Funny. Because there are quite a lot exercices like this and I did them differently getting the right result.
I used binomial distribution with P= (X> 80). Is it wrong? How is that we both come up with the right result?
Thanks a lot

Because They are the same thing.

RonL
• November 4th 2007, 12:35 PM
0123
Quote:

Originally Posted by CaptainBlack
Because They are the same thing.

RonL

oh. :confused: