Question

The Aluminum Association reports that the average American uses 56.8 pounds of aluminum in a year. A random sample of 50 households is monitored for one year to determine aluminum usage. If the population standard deviation of annual usage is 12.2 pounds, what is the probability that the sample mean will be each of the following?

Appendix A Statistical Tables

a. More than 61 pounds
b. More than 56 pounds
c. Between 55 and 57 pounds
d. Less than 54 pounds
e. Less than 49 pounds

(Round the values of z to 2 decimal places. Round your answers to 4 decimal places.)

a. enter the probability that the sample mean will be more than 61 pounds

b. enter the probability that the sample mean will be more than 56 pounds

c. enter the probability that the sample mean will be between 55 and 57 pounds

d. enter the probability that the sample mean will be less than 54 pounds

e. enter the probability that the sample mean will be less than 49 pounds

Answer 1

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^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 of the sampling distribution ( x ) = 56.8
standard Deviation ( sd )= 12.2/ Sqrt ( 50 ) =1.7253
sample size (n) = 50
a.
the probability that the sample mean will be more than 61 pounds
P(X > 61) = (61-56.8)/12.2/ Sqrt ( 50 )
= 4.2/1.725= 2.4343
= P ( Z >2.4343) From Standard Normal Table
= 0.0075
b.
the probability that the sample mean will be more than 56 pounds
P(X > 56) = (56-56.8)/12.2/ Sqrt ( 50 )
= -0.8/1.725= -0.4637
= P ( Z >-0.4637) From Standard Normal Table
= 0.6786
c.
the probability that the sample mean will be between 55 and 57 pounds
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 55) = (55-56.8)/12.2/ Sqrt ( 50 )
= -1.8/1.7253
= -1.0433
= P ( Z <-1.0433) From Standard Normal Table
= 0.1484
P(X < 57) = (57-56.8)/12.2/ Sqrt ( 50 )
= 0.2/1.7253 = 0.1159
= P ( Z <0.1159) From Standard Normal Table
= 0.5461
P(55 < X < 57) = 0.5461-0.1484 = 0.3977
d.
the probability that the sample mean will be less than 54 pounds
P(X < 54) = (54-56.8)/12.2/ Sqrt ( 50 )
= -2.8/1.7253= -1.6229
= P ( Z <-1.6229) From Standard NOrmal Table
= 0.0523
e.
the probability that the sample mean will be less than 49 pounds
P(X < 49) = (49-56.8)/12.2/ Sqrt ( 50 )
= -7.8/1.7253= -4.5208
= P ( Z <-4.5208) From Standard NOrmal Table
= 0