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 57 pounds

c. Between 55 and 57 pounds

d. Less than 55 pounds

e. Less than 49 pounds

Answer 1

This is a normal distribution question with

Sample size (n) = 50
Since we know that

a) P(x > 61.0)=?
The z-score at x = 61.0 is,

z = 2.4344
This implies that
P(x > 61.0) = P(z > 2.4344) = 1 - 0.9925
P(x > 61.0) = 0.0075
b) P(x > 57.0)=?
The z-score at x = 57.0 is,

z = 0.1159
This implies that
P(x > 57.0) = P(z > 0.1159) = 1 - 0.5461
P(x > 57.0) = 0.4539
c) P(55.0 < x < 57.0)=?

This implies that
P(55.0 < x < 57.0) = P(-1.0433 < z < 0.1159) = P(Z < 0.1159) - P(Z < -1.0433)
P(55.0 < x < 57.0) = 0.5461341021515795 - 0.14840468555098485
P(55.0 < x < 57.0) = 0.3977
d) P(x < 55.0)=?
The z-score at x = 55.0 is,

z = -1.0433
This implies that
P(x < 55.0) = P(z < -1.0433) = 0.1484
e) P(x < 49.0)=?
The z-score at x = 49.0 is,

z = -4.521
This implies that
P(x < 49.0) = P(z < -4.521) = 3.0774*10-6
PS: you have to refer z score table to find the final probabilities.
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