Question

The Aluminum Association reports that the average American uses 56.8 pounds of aluminum in a year. A random sample of 51 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 58 pounds
b. More than 57 pounds
c. Between 55 and 58 pounds
d. Less than 55 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 58pounds

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

c. enter the probability that the sample mean will be between 55 and 58pounds

d. enter the probability that the sample mean will be less than 55 pounds

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

Answer 1

Solution :

Given that ,

mean = = 56.8

standard deviation = = 12.4

n = 51

= = 56.8  

= / n = 12.4 / 51 = 1.7363

a) P( > 58 ) = 1 - P( < 58)

= 1 - P[( - ) / < (58 - 56.8) / 1.7363 ]

= 1 - P(z < 0.69 )

= 1 - 0.7549 = 0.2451

Probability = 0.2451

b)

P( > 58 ) = 1 - P( < 57)

= 1 - P[( - ) / < (57 - 56.8) / 1.7363 ]

= 1 - P(z < 0.12 )

= 1 - 0.5478= 0.4522

Probability = 0.4522

c)

P(55< < 58 )  

= P[(55-56.8) /1.7363 < ( - ) / < (58-56.8) /1.7363 )]

= P( -1.04< Z < 0.69 )

= P(Z < 0.69 ) - P(Z < -1.04 )

= 0.7549 - 0.1492 = 0.6057

probability = 0.6057

d)

P( < 55) = P(( - ) / < (55-56.8) /1.7363 )

= P(z < -1.04 )

= 0.1492

probability = 0.1492

e)

P( < 49) = P(( - ) / < (49-56.8) /1.7363 )

= P(z < -4.49 )

= 0

probability = 0.0000