Probability problem??

Question

if the random variable X in the normal distribution(Gaussian distribution) has the average 1(mx=1)and with variance 2 ,find the probability of :

p (-2 <= x <= 1)=??

Answer 1

In the early days, we didn&#039;t have computers. So, once a standard normal distribution was tabulated, we used that. To do this, first normalize the values by finding the z scores:

z = (x - µ) / σ

Since the variance (σ²) is the std deviation squared,

z1 = (-2 - 1) / √2 = -3/2 ∙ √2 ∙ σ

z2 = (1 - 1) / √2 = 0σ

Z2 is 0, is the mean, i.e. 50%. Look up

-2.1213σ, which is something like 1.7%, taking the difference, the answer is about 48.3%. The table is in the back of your textbook. It has a bell shaped curve with a shaded area. It is called a normal curve, or standard curve, or gaussian curve.

use the Area between zero and z table:

http://www.statsoft.com/textbook/sttable...

In XL, use =NORMSDIST(-3/2 ∙ 2^.5) to use the zscore, or

=NORMDIST(-2,1,2^0.5,TRUE) to do it directly, without normalizing

Both return 0.016947

Answer 2

take this part -2&lt;x&lt;1

now subtract  mean and divide by the variance

so it becomes -0.5&lt;z&lt;0

refer the probability tables for p(z&lt;0)-p(z&lt;-0.5)

for the reasons for the above process read any good book like papoulis or peebles etc