In a normal distribution, 4% of the distribution is less than 53 and 97% of the distribution is less than 65.i.Find the mean and standard deviation of the distribution. ii.Find the interquartile range of the distribution.

Answer 1

I have now answered several of these same type of problems for you.
This time show me your steps and your conclusion so I can take a look at it
hint:
What z-score corresponds with P(x < 53) = .04 ?
etc

Answer 2

P(X<53)=0.04
P(X<65)=0.97
53=m-1.751sd----1)
65=m-1.882sd-----2)
Solving this didn't gave me 58.79 which is the answer.

Answer 3

less than 4% ----> z-score = -1.751
less than 97% ---> z-score = 1.881

(53-m)/s = -1.751 ---> m - 1.751s = 53
(65-m)/s = 1.881s = 65 ---> m + 1.881s = 65
subtract them:
3.632s = 12
s = 3.304, then m = 58.785

your 2nd equation was wrong

Answer 4

Can you look the previous question where I added one more question?