Question

3. A) Given that z is a standard normal random variable, compute the probability that it takes on a value between -2 and -1.

3. B). Given that z is a standard normal random variable, find the z-score for a situation where the area to the right of z is 0.0901.

Answer 1

3.A)

Given that , z is a standard normal random variable . we have to find the probability that z lies between -2 and -1 .

So, effectively , we have to find

where, is the cumulative density function of standard normal distribution .

Now , we know that_.

so , our aim reduces to finding_.

now _^.9772

and  _^.8413

Both these values are found in the standard normal table .

so, our required value is (0.9772-0.8413 ) = 0.1359 .

here is the image of areas under normal curve . we use this info to solve the problem .

3.B)

Given that , area to the right of z score is 0.0901 .

we have to find the z score . to solve this , we look into the standard z score table .

we find that area 0.0901 corresponds to -1.3 in the row and 0.04 in the column . so it gives us that the z score is

(-1.3-0.04) = -1.34 .

here is the image required , of the z table , attached .