A random sample of 100 visitors to a popular theme park spent an average of $142 on the trip with a standard deviation of $47.50. Which of the following would be the 98% confidence interval for the mean amount of money spent by all visitors to this theme park?
a. ($130.77, $153.23)
b. ($132.77, $151.43)
c. ($132.69, $151.31)
d. ($140.88, $143.12)

a) 98% confidence interval for the mean amount of money spent by all visitors to this theme park

($130.77, $153.23)

Step-by-step explanation:

Explanation:-

Given sample size 'n' = 100

Given the average of $142 on the trip with a standard deviation of $47.50

mean of the sample (x⁻) =  $142

standard deviation σ = $47.50.

Confidence intervals:-

The 98% of confidence intervals

on calculation, we get

($130.95 , 153.04)

The confidence intervals are ($130.95 , 153.04)

Final -

Its nearly the given option its (a)

($130.77, $153.23)

idk sry

step-by-step explanation:

d. bc = 12 and ef = 16

step-by-step explanation:

we are given that δabc : δdef such that .

also, it is given that

so, from the options, we need that .

a.

b.

c.  

d.

as, we can see that only option d gives the same ratio as i.e. .

hence, the possible lengths are bc = 12 and ef = 16.