(a) The hourly wage of employees in a certain service indust

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(a) The hourly wage of employees in a certain service industry is believed to follow
the Normal Distribution N(40,5^2) which has a mean µ of 40 dollars and a standard
deviation σ of 5 dollars.The hourly wage of an employee is said to be reasonable if it
is between 35 dollars and 45 dollars.A group of 10 employees was selected randomly.
Find the probability that more than 7 employees of the selected group are having a
reasonable hourly wage.
(b) Later a survey was conducted by selecting a group of 64 employees randomly.
The average hourly wage of the selected group of employees is found to be 41 dollars.
Assuming that the standard deviation (σ = 5) remains the same,construct a 95%
confidence interval for the average hourly wage µ

The probability of one employee have reasonable wage is 2φ(1)-1=2*0.8413-1=0.6826.The probability of 8 over 10 employee have reasonable wage is C(10,8)*0.6826^8*(1-0.6826)^2=0.2137.Following the same method,one can get the probability of 9 over 10 employee have reasonable wage is 0.1021 and the probability of all employee have reasonable wage is 0.0220.Therefore,the probability of more than 7 employee have reasonable wage is 0.2137+0.1021+0.0220=0.3378.
The average wage of a group of 64 employee follows the Normal Distribution N(40,5^2/64) with standard derivation 5/8=0.625.Since φ(1.96)=0.975,so the confidence interval is [41-1.96*0.625,41+1.96*0.625] = [39.775,42.225]

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