Answer:
So then we can conclude that we expect the middle 95% of the values within 18 and 30 minutes for this case
Step-by-step explanation:
For this case we can define the random variable X as the amount of time it takes her to arrive to work and we know that the distribution for X is given by:
And we want to use the empirical rule to estimate the middle 95% of her commute times. And the empirical rule states that we have 68% of the values within one deviation from the mean, 95% of the values within two deviations from the mean and 99.7 % of the values within 3 deviations from the mean. And we can find the limits on this way:
So then we can conclude that we expect the middle 95% of the values within 18 and 30 minutes for this case
Answer:
52
Step-by-step explanation 1/3 i believe it would be 1/6 of 314 which would be 52 and 1/3.
So the mean is 72.97
We need to subtract the mean from each value and square it.
(65-72.97)^2= 63.5209
(68-72.97)^2=24.7009
(69-72.97)^2=15.7609
(70-72.97)^2=8.8209
(71-72.97)^2= 3.8809
(72-72.97)^2=0.9409
(90-72.97)^2=290.0209
(95-72.97)^2=485.3209
Now we add up the new values ( also consider their frequency) and find their mean.
Add the values
63.5209+(2 •24.7009=49.4018)+(5•15.7609=78.8045)+(8•8.8209=70.5672)+(7•3.8809=27.1663)+(3•0.9409=2.8227)+(2•290.0209=580.0418)+(2•485.3209=970.6418)= 1,842.967
Divide by total numburs to find the mean
1,842.967/ 30=61.43223333
The standar deviation is the square root of the mean so is
Square root of 61.43223333=7.837871735
Round to the nearest tenth
Standard Deviation is 7.8
21.6 miles per gallon
Multiplied by 6.2 gallons
21.6 * 6.2 = 133.92
Cheryl can travel 133.92 (or about 134) miles on 6.2 gal. of gas.
Hope that helped!
