Answer:
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.
Step-by-step 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 : [tex]P(-1.32 < Z < 1.32)= P(Z<1.32)-P(Z<(-1.32))[/tex]
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
[tex]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\%[/tex]
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.
