Question

suppose a normal distribution has a mean of 79 and a standard deviation of 7 what is p if x is greater than or equal to 93

Answer 1

Given a population X with normal distribution with mean μ=79 and standard deviation σ=7

You have to calculate the probability of X greater than or equal to 93, symbolically:

`P(X\ge93)`

To calculate this probability you have to "transform" this value to the standard normal distribution. You have to use the standard normal or Z distribution because it is tabulated, i.e. its values and accumulated probabilities are known.

To translate a value of any normal distribution to a value of the standard normal distribution you have to subtract the mean and divide by the standard deviation.

The standard normal distribution is defined as:

`Z=(X-\mu)/(\sigma)distN(0,1)`

Calculate the Z value

`Z=(93-79)/(7)=2.00`

So the probability you have to calculate is

`P(X\ge93)=P(Z\ge2.00)`

Now the tables of the standard normal distribution indicate the accumulated probabilities, to determine the probability above a determined value, you have to subtract the probability accumulated until that value to the total probability.

Remember that for any distribution of probability, the total probability is always 1. So under The normal distribution or the standard normal distribution, the total area of under the curve is 1

`P(Z\ge2.00)=1-P(Z<2.00)`

In the Z table (right entry) look for the corresponding value of the probability

`P(Z<2.00)=0.977`

So

`1-P(Z<2.00)=1-0.977=0.023`

The probability of X is greater than or equal to 93 is 0.023

`P(X\ge93)=0.023`