Question

What is the probability that a data value in a normal distribution is between a

z-score of -1.32 and a z score of 1.32? Round your answer to the nearest
tenth of a percent.
A. 84.3%
B. 82.3%
C. 83.3%
D. 81.3%
Please tell me how to do this in my calculator

Answer 1

The probability that a data value in a normal distribution is between a

z-score of -1.32 and a z score of 1.32 is 81.3%

Option D is correct.

Explanation:

We need to find the probability that a data value in a normal distribution is between a z-score of -1.32 and a z score of 1.32.

We can write it as P(-1.32 < Z < 1.32)

It is equal to :
`P(-1.32 < Z < 1.32)= P(Z<1.32)-P(Z<(-1.32))`

Looking at z-score tables to find value of z=1.32 and z=-1.32

P(Z<1.32)= 0.90658

P(Z<-1.32) = 0.09342

Putting values and finding answer

`P(-1.32 < Z < 1.32)= P(Z<1.32)-P(Z<(-1.32))\\ P(-1.32 < Z < 1.32)=0.90658-0.09342\\P(-1.32 < Z < 1.32)=0.81316\\P(-1.32 < Z < 1.32)=81.3\%`

So, the probability that a data value in a normal distribution is between a

z-score of -1.32 and a z score of 1.32 is 81.3%

Option D is correct.