Question

1) Find the​ z-scores that separate the middle 81​% of the distribution from the area in the tails of the standard normal distribution. STEPS ON HOW TO SOLVE IT IN TI83

2) Assume the random variable X is normally distributed with mean u=5050 and standard deviation sigmaσequals=7.

Compute the probability. Be sure to draw a normal curve with the area corresponding to the probability shaded.

​P(56 less than or equals≤X less than or equals≤67​). HOW TO ENTER IT IN TI 83

Answer 1

Solution:

Question 1)

We have to find the​ z-scores that separate the middle 81​% of the distribution from the area in the tails of the standard normal distribution.

Since middle area is 81% = 0.81, we area in tails = 1 - 0.81 = 0.19

We divide this 0.19 area in two tails equally, that is: 0.19/2=0.095 in left tail and 0.095 in right tail.

We use 0.095 area to find z value in left tail.

Use following steps in TI 83 / 84 plus calculator.

Press 2ND and VARS

Select invNorm(

Enter numbers:

Click on Paste and press enter two times

Thus z value for left tail is -1.31

Now use Area = 0.095+0.81 = 0.905 to find z value in right tail

use same steps above

Enter Area as 0.905

Click on Paste and press enter two times

Thus z value = 1.31

Thus z values that separate the middle 81​% of the distribution from the area in the tails of the standard normal distribution are: (-1.31 , 1.31 )

Question 2)

the random variable X is normally distributed with mean and standard deviation  .

Use following steps in TI 83:

Press 2ND and VARS

select normalcdf(

Enter numbers:

Click on Paste and press enter two times

thus