Answer: The standard error for sample is 0.0686 .
Step-by-step explanation:
We know that the formula to find the standard error is given by :-
, where s = standard deviation
n= Sample size
As per given , we have
s= $0.4 and n= 34
Then , the standard error for sample is given by :-
Hence , the standard error for sample is 0.0686 .
Answer:
Probability that the average mileage of the fleet is greater than 30.7 mpg is 0.7454.
Step-by-step explanation:
We are given that a certain car model has a mean gas mileage of 31 miles per gallon (mpg) with a standard deviation 3 mpg.
A pizza delivery company buys 43 of these cars.
<em>Let </em><em> = sample average mileage of the fleet </em>
<em />
The z-score probability distribution of sample average is given by;
Z = ~ N(0,1)
where, = mean gas mileage = 31 miles per gallon (mpg)
= standard deviation = 3 mpg
n = sample of cars = 43
So, probability that the average mileage of the fleet is greater than 30.7 mpg is given by = P(<em> </em>> 30.7 mpg)
P(<em> </em>> 30.7 mpg) = P(
>
) = P(Z > -0.66) = P(Z < 0.66)
= 0.7454
<em>Because in z table area of P(Z > -x) is same as area of P(Z < x). Also, the above probability is calculated using z table by looking at value of x = 0.66 in the z table which have an area of 0.7454. </em>
Therefore, probability that the average mileage of the fleet is greater than 30.7 mpg is 0.7454.