Have I calculated the probability of these z-scores correctly?

Question

For a normal distribution, use the given z-scores to find the requested probability:

Q1: Find the probability greater than or equal to a z-score of 2

A1: P(z>2) = 1 - p(z<2) = 1 - .9972 = .0228 or 2.28%

 

Q2: Find the probability less than or equal a z-score of -2

A2: p(p<-2) = .0228 or 2.28%

 

Q3: Find the probability less than or equal to a z-score of 2

A3: p(z<2) = .9972 or 97.72%

 

Q4: Find the probability between z-scores of -1.50 and 2

A4: p(-1.50 < z < 2) = .0668 + .9772 = 1.044 or 104.4%

 

I do not feel right about the last question especially since a probability shouldn't exceed 100%.

Please help Thank you

Answer 1

You got the first three correct. I verified the calculations using R statistical software:

> 1-pnorm(2)
[1] 0.02275013
> pnorm(-2)
[1] 0.02275013
> pnorm(2)
[1] 0.9772499
> pnorm(2)-pnorm(-1.5)
[1] 0.9104427

You are correct that a probability should never exceed $1$. For problem $4$ you need to find the probability of being $between$ $-1.5$ and $2$. So you need to find the probability of being less than $2$ and subtract the probability of being less than $-1.5$. We have,

$$\begin{align*} P(-1.5<Z<2) &=P(Z<2)-P(Z<-1.5)\\\\ &\approx0.9772-0.0668\\\\ &\approx0.9104 \end{align*}$$

Answer 2

Next time, if it helps, try to visualize it!

We're talking about what the probability is that a random observation has a given z-score. $P(Z<2)$ corresponds to everything blue (including the green overlap). $P(Z<-1.5)$ corresponds to the green part. So $P(-1.5<Z<2)$ is the only blue part, which we can evidently see is $P(Z<2)-P(Z<-1.5)$