Answer to Question #196496 in Statistics and Probability for nahal zehra

Question

It is known that the monthly income of salaried persons in a large city are normally distributed with a standard deviation of 3600 Rs.. (a) A random sample of 9 individuals from this population is taken and it is found that their salaries are as follows: 44000, 41000, 43000, 42500, 43500, 41500, 41000, 46500, 35000 What is the probability that the average salary of the whole salaried population in the city lies somewhere between 40025 and 43975? 

Answer 1

We have given that,

$\sigma = 3600$

$\mu = \dfrac{44000+ 41000+ 43000+ 42500+ 43500+ 41500+ 41000+ 46500+ 35000}{9}$

$\mu = 42000$

We have to find the probability that the average salary of the whole salaried population in the city lies somewhere between 40025 and 43975

$P(40025<X<43975)$

$= P(\dfrac{40025-42000}{3600}<Z<\dfrac{43975-42000}{3600})$

$= P(-0.54<Z<0.54)\\ = 2P(0<Z<0.54)\\ = 0.41$