I found the picture of the bowl.

AB is the diameter of the sphere.

So the radius is 20in / 2 = 10 in

And the volume of water is half the volume of the sphere.

=> V of water = (1/2)* (4/3) π (r^3)

Also, this problems asks to use 22/7 as approximation of π.

=> V of water = (1/2) (4/3) (22/7) (10in)^3 = 2,095 in^3

Answer: (1/2)(4/3)(22/7)(10in)^3 = 2,095 in^3

3 0

3 years ago

To estimate the mean height μ of male students on your campus, you will measure an SRS of students. You know from government dat

denis-greek [22]

Answer:

a) The standard deviation of x must be 0.25 inches.

b) We need a sample of 125 to reduce the standard deviation of x⎯⎯⎯ to the value you found in part(a).

Step-by-step explanation:

The 68-95-99.7 states that:

68% percent of the measures of a normally distributed sample are within 1 standard deviation of the mean.

95% percent of the measures of a normally distributed sample are within 2 standard deviations of the mean.

99.7% percent of the measures of a normally distributed sample are within 3 standard deviations of the mean.

The standard deviation of the population is 2.8. This means that $\sigma = 2.8$ .

(a) What standard deviation must x⎯⎯⎯ have so that 95% of all samples give an x⎯⎯⎯ within one-half inch of μ?

We want to have a sample in which 2 standard deviations are within 0.5 inches of the mean.

So, the standard deviation of the sample must be:

$2s = 0.5$

$s = 0.25$

The standard deviation of x must be 0.25 inches.

(b) How large an SRS do you need to reduce the standard deviation of x⎯⎯⎯ to the value you found in part(a)?

We have that the standard deviation of a sample of length n is given by the following formula:

$s = \frac{\sigma}{\sqrt{n}}$ .

We want $s = 0.25$ and we have $\sigma = 2.8$ . So

$0.25 = \frac{2.8}{\sqrt{n}}$

$0.25\sqrt{n} = 2.8$

$\sqrt{n} = 11.2$

$\sqrt{n}^{2} = (11.2)^{2}$

$n = 125.44$

We need a sample of 125 to reduce the standard deviation of x⎯⎯⎯ to the value you found in part(a).

8 0

4 years ago