SOLUTION: for a normal distribution if the value of mean is 70 and standard deviation is 15.75 find the points (a) in which it contains 98% area between them (b) in which it contains 95% a

Question 1191928: for a normal distribution if the value of mean is 70 and standard deviation is 15.75 find the points
(a) in which it contains 98% area between them
(b) in which it contains 95% area between them
(c) in which it contains 85% area between them
(d) in which 60% area between them

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Part (a)

Your teacher is asking you to find the values of m and n such that
P(m < X < n) = 0.98
The problem is that there are infinitely many solutions for m,n.

It's similar to how
P(r < Z < s) = 0.98
also has infinitely many solutions for r,s

However, if we were to use symmetry and say
P(-k < Z < k) = 0.98
then there's exactly one solution for k

For the duration of this solution, I'll assume that your teacher is using the P(-k < Z < k) method.

98% of the area is in the main body leaving 100% - 98% = 2% in the two tails combined.
Each tail would have (2%)/2 = 1% of the area.
We're able to divide by two because of the mirror symmetry mentioned.

Use a table or calculator to find that P(Z < k) = 0.01 has a solution of roughly k = -2.327
Here's a useful calculator
https://davidmlane.com/normal.html

That k value leads to
P(-2.327 < Z < 2.327) = 0.98
Meaning that about 98% of the area under the standard normal Z curve is between the z scores -2.327 and 2.327

Now let's convert each mentioned z score into their corresponding raw score.

We'll start with z = -2.327
z = (x-mu)/sigma
z*sigma = x-mu
x = mu+z*sigma
x = 70+(-2.327)*15.75
x = 33.34975

Then repeat for z = 2.327
x = mu+z*sigma
x = 70+2.327*15.75
x = 106.65025

Therefore,
P(33.34975 < X < 106.65025) = 0.98
approximately

You'll follow similar steps for parts (b) through (d).

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