Ahotel chain wants to estimate the average number of rooms rented daily in each month. the population of rooms rented daily is assumed to be normally distributed for each month with a standard deviation of 24 rooms. during june, a sample of 16 days has a sample mean of 48 rooms. this information is used to calculated an interval estimate for the population mean to be from 40 to 56 rooms. what is the level of confidence of this interval?

81.64%

Step-by-step explanation:

Data provided in the question:

Standard deviation, s = 24 rooms

Sample size, n = 16 days

Mean = 48 rooms

calculated an interval = 40 to 56

Now,

Level of confidence = 100(1 - α)%

Confidence interval = Mean ±

Thus,

Mean - = 40

and,

Mean + = 56

or

48 + = 56

or

= 8

or

z = 1.33

for z = 1.33 , we have = 0.918  [from standard z table]

or

α = 0.1836

Therefore,

Level of confidence = 100(1 - 0.1836)%

or

Level of confidence = 81.64%

central angle of a triangle has the same measure as its intercepted arc.

step-by-step explanation:

0 and 5 minutes

step-by-step explanation:

given that one morning eight buses arrive late by the number of minutes given below:

0 8 2 6 9 2 0 7 8

write these in ascending order

0   0 2 6 7 8 8 9

in the afternoon two more buses arrive at the bus stop say by x and y minutes late.

total number of buses = 10

median = average of 4th and 5th after arranging in ascending order

since 4th entry is 6, the values of x and y should be below 6

also given 0 is the mode so either x or y or both should be 0 because then only 0 gets the maximum frequency and hence mode

thus we have

0 0 0 2 6 7 8 8 9 and another unknown entry which is below 6.

since median is 3.5, we have the unknown entry to be 5 only because then only median =( 2+5)/2

thus the after noon buses were late by 0 and 5 minutes.