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?

a. More than 61 pounds

b. More than 56 pounds

c. Between 56 and 58 pounds

d. Less than 54 pounds

e. Less than 49 pounds

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

Answer 1

Answer:

Given,

Mean = 56.8

Standard deviation = 12.2

Sample = 50

a)

P(X > 61) = P((x-u)/(s/sqrt(n)) > (61 - 56.8)/(12.2/sqrt(50)))

= P(z > 2.43)

= 0.0075494 [since from z table]

= 0.0075

b)

P(X > 56) = P((x-u)/(s/sqrt(n)) > (56 - 56.8)/(12.2/sqrt(50)))

= P(z > -0.46)

= 0.6772419 [since from z table]

= 0.6772

c)

P(56 < X < 58) = P((56 - 56.8)/(12.2/sqrt(50)) < (x-u)/(s/sqrt(n)) < (58 - 56.8)/(12.2/sqrt(50)))

= P(-0.46 < z < 0.70)

= P(z < 0.70) - P(z < -0.46)

= 0.7580363 - 0.3227581 [since from z table]

= 0.4353

d)

P(X < 54) = P((x-u)/(s/sqrt(n)) < (54 - 56.8)/(12.2/sqrt(50)))

= P(z < -1.62)

= 0.0526161 [since from z table]

= 0.0526

e)

P(X < 49) = P((x-u)/(s/sqrt(n)) < (49 - 56.8)/(12.2/sqrt(50)))

= P(z < -4.52)

= 0.0000031 [since from z table]

= 0.0000