# Thread: Probability calculations using normal distribution.

1. ## Probability calculations using normal distribution.

The life-time of a popular computer flatscreen is known to follow a normal distribution with a mean μ = 2000 hours and standard deviation σ = 200 hours. If a computer flatscreen is selected at random, find the probability that it will last for:
Less than 2500 hours.
More than 1900 hours.
Between 2000 and 2400 hours.

can someone guide me tru this problem

2. For example

$p(x\leq 2500)=p(z\leq (2500-2000)/200)=\ldots$

Now, use the table of the standardised normal distribution $N(0,1)$ .

3. Originally Posted by math321
The life-time of a popular computer flatscreen is known to follow a normal distribution with a mean μ = 2000 hours and standard deviation σ = 200 hours. If a computer flatscreen is selected at random, find the probability that it will last for:
Less than 2500 hours.
More than 1900 hours.
Between 2000 and 2400 hours.

can someone guide me tru this problem
If you need further help, please show your working and say where you are stuck. And since this type of question is so routine, you would clearly benefit from reviewing examples from your class notes and textbook.

4. so after i get to

$p(z\leq 2.5$
wat do i do from there

got 0.4938 on the table for 2.5

5. Originally Posted by math321
so after i get to

$p(z\leq 2.5$
wat do i do from there

got 0.4938 on the table for 2.5
This is clearly wrong. You should know that Pr(Z < 0) = 0.5, so obviously Pr(Z < 2.5) must be greater than 0.5. You need to get one-on-one help from your instructor on how to use the tables you are using. It is the same sort of question as the ones you posted here: http://www.mathhelpforum.com/math-he...on-176941.html.

6. can u just break down the question into steps so i can understand it better

if that is wrong wat do i do from there

7. Originally Posted by math321
can u just break down the question into steps so i can understand it better

if that is wrong wat do i do from there
Remember one thing, the table values are for
$P(Z>z)$ where $z$ is your specific $Z$ value.

8. stuck here

Z = (2500 - 2000) / 200

z = 2.5

9. is it this
P( X < 2500)
= P(z < 2.5)
= .9953

10. Originally Posted by math321
is it this
P( X < 2500)
= P(z < 2.5)
= .9953

No, it is .9938 because $p(z\leq a)=p(z .

11. and for the 2nd part of the question
z = (-100 / 200)

z = -0.5
=P(Z > -.5)
=P(Z < .5)

= .6915

1- .6915

12. Originally Posted by math321
=P(Z > -.5)
=P(Z < .5)
= .6915
1- .6915

No, it is

$p(z>-0.5)=p(z<0.5)=0.6915$

13. Originally Posted by math321
can u just break down the question into steps so i can understand it better

if that is wrong wat do i do from there
You are asking the same sort of questions every time in all your threads and getting given the same help. It is time you went back and reviewed the examples in your textbook and class notes and got one-on-one help in person from someone (eg. your instructor).