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A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 27 weeks. Assume that the length of unemployment is normally distributed with population mean of 27 weeks and the population standard deviation of 2 weeks. Suppose you would like to select a random sample of 39 unemployed individuals for a follow-up study.

Required:
a. What is the distribution of X?
b. What is the distribution of xÌ?
c. What is the probability that d. For 36 unemployed individuals, find the probability that the average time that they found the next job is less than one randomly selected individual found a job less than 27 weeks?

Sagot :

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Answer:

X ~ N(27, 4) ;

xbar ~ N(27, 0.1026) ;

0.5 ;

0.5

Step-by-step explanation:

Probability distribution of X : N(μ, σ²)

μ = 27 ; σ = 2

X ~ N(μ, σ²) = X ~ N(27, 2²) ;X ~ N(27, 4)

Distribution is approximately normal ; μ = xbar ; xbar = 27

(Standard Error)² = (σ/√n)²= (2/√39)² = 0.1026

xbar ~ N(μ, σ²) = xbar ~ N(27, 2²) ; xbar ~ N(27, 0.1026)

Probability that a randomly selected individual found a job in less than 27 weeks :

P(X < 27) :

Obtain the Zscore :

Z = (x - μ) / σ

Z = (27 - 27) / 2 = 0/2

Z = 0

P(Z < 0) = 0.5

D.) n = 36

P(X < 27) :

Obtain the Zscore :

Z = (x - μ) / σ/√n

Z = (27 - 27) / (2/√36) = 0/0.33333

Z = 0

P(Z < 0) = 0.5

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