Mathematics, 25.07.2020 04:01 layahmonaee

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 12.2 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 58 pounds d. Less than 55 pounds e. Less than 48 pounds

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Answer from: tamikagoss22

(a) The probability that the sample mean will be more than 61 pounds is 0.0069.

(b) The probability that the sample mean will be more than 57 pounds is 0.4522.

(d) The probability that the sample mean will be less than 55 pounds is 0.14686.

(e) The probability that the sample mean will be less than 48 pounds is 0.00001.

Step-by-step explanation:

We are given that 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. Also, the population standard deviation of annual usage is 12.2 pounds.

Let $\bar X$ = sample mean

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = average aluminum used by American = 56.8 pounds

$\sigma$ = population standard deviation = 12.2 pounds

n = sample of households = 51

(a) The probability that the sample mean will be more than 61 pounds is given by = P( $\bar X$ > 61 pounds)

P( $\bar X$ > 61 pounds) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{61-56.8}{\frac{12.2}{\sqrt{51} } }$ ) = P(Z > 2.46) = 1 - P(Z $\leq$ 2.46)

= 1 - 0.9931 = 0.0069

The above probability is calculated by looking at the value of x = 2.46 in the z table which has an area of 0.9931.

(b) The probability that the sample mean will be more than 57 pounds is given by = P( $\bar X$ > 57 pounds)

P( $\bar X$ > 57 pounds) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{57-56.8}{\frac{12.2}{\sqrt{51} } }$ ) = P(Z > 0.12) = 1 - P(Z $\leq$ 0.12)

= 1 - 0.5478 = 0.4522

The above probability is calculated by looking at the value of x = 0.12 in the z table which has an area of 0.5478.

(c) The probability that the sample mean will be between 55 and 58 pounds is given by = P(55 pounds < $\bar X$ < 58 pounds)

P(55 pounds < $\bar X$ < 58 pounds) = P( $\bar X$ < 58 pounds) - P( $\bar X$ $\leq$ 55 pounds)

P( $\bar X$ < 58 pounds) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{58-56.8}{\frac{12.2}{\sqrt{51} } }$ ) = P(Z < 0.70) = 0.75804

P( $\bar X$ $\leq$ 55 pounds) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{55-56.8}{\frac{12.2}{\sqrt{51} } }$ ) = P(Z $\leq$ -1.05) = 1 - P(Z < 1.05)

= 1 - 0.85314 = 0.14686

The above probability is calculated by looking at the value of x = 0.70 and x = 1.05 in the z table which has an area of 0.75804 and 0.85314.

Therefore, P(55 pounds < $\bar X$ < 58 pounds) = 0.75804 - 0.14686 = 0.6112.

(d) The probability that the sample mean will be less than 55 pounds is given by = P( $\bar X$ < 55 pounds)

P( $\bar X$ < 55 pounds) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{55-56.8}{\frac{12.2}{\sqrt{51} } }$ ) = P(Z < -1.05) = 1 - P(Z $\leq$ 1.05)

= 1 - 0.85314 = 0.14686

The above probability is calculated by looking at the value of x = 1.05 in the z table which has an area of 0.85314.

(e) The probability that the sample mean will be less than 48 pounds is given by = P( $\bar X$ < 48 pounds)

P( $\bar X$ < 48 pounds) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{48-56.8}{\frac{12.2}{\sqrt{51} } }$ ) = P(Z < -5.15) = 1 - P(Z $\leq$ 5.15)