MK Mary Kay iuww w1NS D2L e chegg D wileypus LAPA citation generat Other bookma

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MK Mary Kay iuww w1NS D2L e chegg D wileypus LAPA citation generat Other bookma Black, Business Statistics, 8e BUSINESS STATISTICS (ECON 245) ractice Assignment Gradebook ORION gnment 4BACK NEXT The U.S. Bureau of Labor Statistics released hourly wage figures for various countries for workers in the manufacturing sector. The hourly wage was $30.67 for Switzerland, $20.20 for Japan, and $23.82 for the U.S. Assume that in all three countries, the standard deviation of hourly labor rates is $4.00. a. Suppose 42 manufacturing workers are selected randomly from across Switzerland and asked what their hourly wage is. What is the probability that the sample average will be between $30.00 and $31.007 b. Suppose 38 manufecturing workers are selected randomly from across Japan. What is the probability that the sample average will exceed $21.00? C. Suppose 47 manufacturing workers are selected randomly from across the United States. What is the probability that the sample average will be less than $22.85? fRound the values of z to 2 decimal places. Round your answers to 4 decimal places.) b. Question Attempts: 0 of 2 used SAVE FOR LATER SUBMIT ANSWER n Ios Al Rights Raserved. A Divis

Explanation / Answer

a.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean ( u ) = 30.67
standard Deviation ( sd )= 4/ Sqrt ( 42 ) =0.6172
sample size (n) = 42
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 30) = (30-30.67)/4/ Sqrt ( 42 )
= -0.67/0.617213
= -1.085524
= P ( Z <-1.085524) From Standard Normal Table
= 0.138845
P(X < 31) = (31-30.67)/4/ Sqrt ( 42 )
= 0.33/0.617213 = 0.534661
= P ( Z <0.534661) From Standard Normal Table
= 0.703558
P(30 < X < 31) = 0.703558-0.138845 = 0.564713

b.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean ( u ) = 20
standard Deviation ( sd )= 4/ Sqrt ( 38 ) =0.6489
sample size (n) = 38
P(X > 21) = (21-20)/4/ Sqrt ( 38 )
= 1/0.649= 1.5411
= P ( Z >1.5411) From Standard Normal Table
= 0.0616

c.

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean ( u ) = 23.82
standard Deviation ( sd )= 4/ Sqrt ( 47 ) =0.5835
sample size (n) = 47
P(X < 22.85) = (22.85-23.82)/4/ Sqrt ( 47 )
= -0.97/0.5835= -1.6625
= P ( Z <-1.6625) From Standard NOrmal Table
= 0.048218

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