Probability 3+4 Question? HELP!?? D:

[john.wu92]

I'm stuck on this Normal Distribution question, its killing me D;

The Weekly error of a watch (in seconds) is known to be normally distributed. Only those watches with an error of less than 5 seconds are accepted.
a) find the mean and standard deviation of the distribution of error if 3% are rejected for losing time and 3% are rejected for gaining time.


Thanks!

[taiga]

The mean is 0 because it is a normal distribution with a rejection region of equal magnitude from 0. (let me know if you need more explaining on this one)

Using a standard normal distribution, find the point Z which corresponds to the point with 0.97 of the data before it. That value is 1.880793608 I believe.



We use the above Z value, we use +5 as our X value, 0 as the mean, and we want to find the standard deviation.

hence

1.880793608 = 5/standard dev

Standard Dev = 5/1.880793608

[john.wu92]

Using a standard normal distribution, find the point Z which corresponds to the point with 0.97 of the data before it. That value is 1.880793608 I believe.

Thanks, the answer is right, however, How did You get 1.88? :O

Thats what I don't get

Thank You.

[taiga]

So if we are converting to Z, we know we are converting to a mean of 0, and standard deviation of 1.

There is calculator syntax (I used microsoft excel's NORMSINV), I think it is INVNORM on the TI-Nspire, and that should spit out the answer you require.

It is very important to know this sort of syntax come exam 2!