Question

The Aluminum Association reports that the average American uses 56.8 pounds of aluminum in a year. A random sample of 49 households is monitored for one year to determine aluminum usage. If the population standard deviation of annual usage is 12.4 pounds, what is the probability that the sample mean will be each of the following?

Appendix A Statistical Tables

a. More than 60 pounds
b. More than 57 pounds
c. Between 56 and 57 pounds
d. Less than 54 pounds
e. Less than 48 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 60 pounds

b. enter the probability that the sample mean will be more than 57 pounds

c. enter the probability that the sample mean will be between 56 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 48 pounds

Answer 1

Solution :

Given that,

mean = = 56.8

standard deviation = = 12.4

n = 49

= 56.8

₌ / n = 12.4 49 = 1.7714

a ) P (x > 60)

= 1 - P ( < 60 )

= 1 - P ( - / ) < ( 60 - 56.8 / 1.7714)

= 1 - P ( z < 3.2 / 1.7714 )

= 1 - P ( z < 1.81)

Using z table

= 1 - 0.9649

= 0.0351

Probability = 0.0351

b ) P (x > 57)

= 1 - P ( < 57)

= 1 - P ( - / ) < ( 57 - 56.8 / 1.7714)

= 1 - P ( z < - 0.2 / 1.7714 )

= 1 - P ( z < 0.11)

Using z table

= 1 - 0.5438

= 0.4562

Probability = 0.4562

c ) P (56 < < 57 )

P ( 56 - 56.8 / 1.7714) < ( - / ) < ( 57 - 56.8 /1.7714)

P ( - 0.8 / 1.7714 < z < 0.2 / 1.7714)

P (-0.45 < z < 0.11 )

P ( z < 0.11 ) - P ( z < -0.45)

Using z table

= 0.5438 - 0.3264

= 0.2174

Probability = 0.2174

d ) P (   < 54 )

P ( - /) < (54 - 56.8 / 1.7714)

P ( z < -2.8 / 1.7714 )

P ( z < -1.58 )

Using z table

= 0.0571

Probability = 0.0571

e ) P (   < 48 )

P ( - /) < (48 - 56.8 / 1.7714)

P ( z < -8.8 / 1.7714 )

P ( z < - 4.97 )

Using z table

= 0

Probability = 0