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
.
(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](https://tex.z-dn.net/?f=2s%20%3D%200.5)
![s = 0.25](https://tex.z-dn.net/?f=s%20%3D%200.25)
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:
.
We want
and we have
. So
![0.25 = \frac{2.8}{\sqrt{n}}](https://tex.z-dn.net/?f=0.25%20%3D%20%5Cfrac%7B2.8%7D%7B%5Csqrt%7Bn%7D%7D)
![0.25\sqrt{n} = 2.8](https://tex.z-dn.net/?f=0.25%5Csqrt%7Bn%7D%20%3D%202.8)
![\sqrt{n} = 11.2](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%2011.2)
![\sqrt{n}^{2} = (11.2)^{2}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%5E%7B2%7D%20%3D%20%2811.2%29%5E%7B2%7D)
![n = 125.44](https://tex.z-dn.net/?f=n%20%3D%20125.44)
We need a sample of 125 to reduce the standard deviation of x⎯⎯⎯ to the value you found in part(a).