Consider the following probability.

Question

Consider the following probability.

P(−1.24 ≤ z ≤ 2.73)

Let z be a random variable with a standard normal distribution. Find the indicated probability. (Round your answer to four decimal places.)

P(−1.24 ≤ z ≤ 2.73) =

Answer 1

.39251 for z= -1.24 to 0

.49683 for z= 0 to 2.73

add them together

.39261 + .49683 = .88944

= about .8894

use z tables or a z calculator

Answer 2

Hi Tamera,

This is our old friend: z= (x-mu)/sigma. Now, in a standard normal distribution, mean is 0 and standard deviation is 1, so we have our mu and sigma, and we were asked to find probability of a range between two x values. We need z-scores to do this, so look up the z table online. Since we need two z-values here, I will call the first z z1 and the second z2. Same with the x values.

z₁ = x₁ - mu / sigma

x₁ = -1.24

mu=0

sigma=1

z₁= -1.24

z₂ = x₂ -mu / sigma

z₂ = (2.73 - 0) /1

z₂ = 2.73

If you know the data is standard normal, you do not even need to do those calculations. You can assume x=z and go right to the z-table. But remember, z probabilities are less than, so:

P(z₁ < -1.24) = 0.1075

P(z₂ < 2.73) = 0.9968

Now, a <= b <= c implies the area between. To get that, we subtract:

P = 0.9968 - 0.1075 = 0.8893

I hope this helps. I noticed you have a few problems on here relating to the normal distribution. Contact me on Wyzant if you want to schedule a live tutoring session and we can run through some more examples.