standard deviation, the probability is 0.20 and value greater than 79.2. normal distribution with mean of 62.4.

Well, what you know:

X \sim N(62.4, \sigma ^2)

Then, P(X>79.2) = 0.2

1. Find the z value that corresponds to this probability.

2. Use z = \frac{x - \mu}{\sigma} to find the value of the standard deviation.

Post what you get! :)

When using the z score equation and you know the mean and standard deviation, but how do you figure out the X. Is x the mode or median?

when you have the probability and the mean, how do u figure out the standard deviation and the z score. Mean is 47.3. probability is 0.25.

You've posted this question like 4 times now. Could you please stop as it just confuses everyone.

I'll give you an example. Try to work your questions and post what you did before asking a question next time.

Qu. The heights of female students at a particular college are normally distributed with a mean of 169 cm and a standard deviation of 9 cm.
(a) Given that 80% of these female students have a height less than h cm, find the value of h.
(b) Given that 60% of these female students have a height greater than s cm, find the value of s.

Solution:
Let X represent the height, in centimetres, of a female student.

X \sim N(169, 9^2)

(a) Given P(X< h) = 0.8

Standardising;

P\(Z < \frac{h - 169}{9}\) = 0.8

Let z = \frac{h - 169}{9}

P(Z<z) = 0.8

The corresponding z value in your table for this probability is 0.842

Hence,

0.842 = \frac{h - 169}{9}

So, h = 0.842(9) + 169 = 176.38 \approx 176.4\ (1\ d.p.)

(b) Given P(X >s) = 0.6

Standardising;

P\(Z > \frac{s - 169}{9}\) = 0.6

Let z = \frac{s - 169}{9}

P(Z>z) = 0.6

But if you made a sketch of the normal table and shaded the right area, you'll see that z must be negative.

This z value corresponding to the probability is given by -0.253 in your table.

Hence

-0.253 = \frac{s - 169}{9}

So, s = -0.253(9) + 169 = 166.723 \approx 166.7\ (1\ d.p.)

Questions? :)