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 57 pounds

c. Between 55 and 57 pounds

d. Less than 55 pounds

e. Less than 47 pounds

Answer 1

Solution :

Given that ,

$\mu$
_T
= 56.8

$\sigma$
_T
= $\sigma$ / $\sqrt$ n = 12.2 / $\sqrt$ 50

a) P(
T
> 61) = 1 - P(
T
< 61)

= 1 - P[(
T
- $\mu$
_T
) / $\sigma$
_T
< (61 - 56.8) / 12.2 / $\sqrt$ 50 ]

= 1 - P(z < 2.43)   

= 1 - 0.9925

= 0.0075

b) P(
T
> 57) = 1 - P(
T
< 57)

= 1 - P[(
T
- $\mu$
_T
) / $\sigma$
_T
< (57 - 56.8) / 12.2 / $\sqrt$ 50 ]

= 1 - P(z < 0.12)   

= 1 - 0.5478

= 0.4522

c) P(55 < < 57 )  

T

= P[(55 - 56.8) / 12.2 / $\sqrt$ 50 < (
T
- $\mu$
_T
) / $\sigma$
_T
< (57 - 56.8) /12.2 / $\sqrt$ 50)]

= P(-1.04 < Z < 0.12)

= P(Z < 0.12) - P(Z < -1.04)

Using z table,  

= 0.5478 - 0.1492

= 0.3986

d) P(
T
< 55) = P((
T
- $\mu$
_T
) / $\sigma$
_T
< (55 - 56.8) / 12.2 / $\sqrt$ 50)

= P(z < -1.04)

Using z table

= 0.1492

e) P(
T
< 47) = P((
T
- $\mu$
_T
) / $\sigma$
_T
< (47 - 56.8) / 12.2 / $\sqrt$ 50)

= P(z < -5.68

Using z table

= 0