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patriot [66]
3 years ago
9

Comparing observations from different populations: The heights of adult men in America are normally distributed, with a mean of

69.7 inches and a standard deviation of 2.66 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.5 inches and a standard deviation of 2.54 inches.
Required:
a. If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)?
b. What percentage of men are SHORTER than 6 feet 3 inches?
c. If a woman is 5 feet 11 inches tall, what is her z-score (to two decimal places)?
d. What percentage of women are TALLER than 5 feet 11 inches?
Mathematics
1 answer:
Schach [20]3 years ago
3 0

Answer:

a. Z = 1.99

b. 97.67%

c. Z = 2.56

d. 0.52%

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

a. If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)?

The heights of adult men in America are normally distributed, with a mean of 69.7 inches and a standard deviation of 2.66 inches, and thus, we have \mu = 69.7, \sigma = 2.66

6 feet 3 inches = 6*12 + 3 = 75 inches, which means that we have to find z when X = 75. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{75 - 69.7}{2.66}

Z = 1.99

b. What percentage of men are SHORTER than 6 feet 3 inches?

The proportion is the p-value of Z = 1.99.

Looking at the z-table, Z = 1.99 has a p-value of 0.9767.

0.9767*100% = 97.67%, which is the answer.

c. If a woman is 5 feet 11 inches tall, what is her z-score (to two decimal places)?

The heights of adult women in America are also normally distributed, but with a mean of 64.5 inches and a standard deviation of 2.54 inches, and thus, we have \mu = 64.5, \sigma = 2.54. We have to find Z when X = 5*12 + 11 = 71. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{71 - 64.5}{2.54}

Z = 2.56

d. What percentage of women are TALLER than 5 feet 11 inches?

The proportion is 1 subtracted by the p-value of Z = 2.56.

Looking at the z-table, Z = 2.56 has a p-value of 0.9948.

1 - 0.9948 = 0.0052

0.0052*100% = 0.52%, which is the answer.

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The sum of the radius of the base and the height of a solid cylinder is 37m. If the total surface area of the cylinder is 1628 m
MA_775_DIABLO [31]

Answer:

4618.14m^{3} or 4618m^{3}  

Step-by-step explanation:

From  T.S.A = 2\pirh + 2\pir^{2}

 where T.S.A = 1628m^{2}

           1628 = 2\pirh + 2\pir^{2}

            1628 =  ( rh + r^{2} ) 2\pi

 by dividing both side by 2\pi

             \frac{1628}{2\pi }=  rh + r^{2}

          259.10 = rh + r^{2}

           rh = 259.10 - r^{2}

        h = \frac{259.10 - r^{2} }{r}   <u>          </u>(1)

From Radius + Height = 37

       r + h = 37  <u>              </u>(2)

by substituting eqn 1 into 2

    r + \frac{259.10 - r^{2} }{r}   = 37

by multiplying r to both side

   r^{2} + 259.10 - r^{2} = 37r

     259.10 = 37r

   r = \frac{259.10}{37}

   r = 7.00 ≅ 7  

From Eqn 2

     r + h = 37

     7 + h = 37

     h = 37 - 7

     h = 30

so Volume of a cylinder = \pi r^{2} h

                  V = \pi * 7^{2} * 30

                  V = 4618.14m^{3}  ≅  4618m^{3}  

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