find probability

Question

A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 30 weeks. Assume that the length of unemployment is normally distributed with population mean of 30 weeks and the population standard deviation of 7 weeks. Suppose you would like to select a random sample of 38 unemployed individuals for a follow-up study. Round the answers of following questions to 4 decimal places.

1. What is the probability that one randomly selected individual found a job more than 31 weeks? 
2. For 38 unemployed individuals, find the probability that the average time that they found the next job is more than 31 weeks. 

Answer 1

Using the TI-83, 83+, 84, 84+ Calculator to calculate these probabilities

Go to 2nd DISTR, and select item 2: normalcdf

The syntax is: normalcdf (lower bound, upper bound, mean, standard deviation)

1)

[population] X follows a normal distribution, with mean = 30, and standard deviation of 7

That means X ~ N (30, 7)    

P (X > 31)

= normalcdf (-1E99, 31, 30, 7)

= 0.5568

2)

[random sample of 38] X bar, follows a normal distribution, with mean = 30, and std dev of 7/sqrt(38) That means X bar ~ N (30, 7/sqrt(38))

P (X bar > 31)

= normalcdf (-1E99, 31, 30, 7/sqrt(38))

= 0.8107