A car rental company is interested in the amount of time its vehicles are out of operation for repair work. A random sample of 21 cars showed that, over the past year, the numbers of days that each of these cars had been inoperative were as follows: 15, 11, 19, 24, 6, 18, 20, 15, 18, 12, 14, 19, 5, 21, 12, 11, 8, 10, 7, 2, 14 Assuming that the population is normally distributed, construct a 95% confidence interval for the mean number of days that vehicles in the company's fleet are out of operation.

(10.891; 15.869)

Step-by-step explanation:

Given the data::

15, 11, 19, 24, 6, 18, 20, 15, 18, 12, 14, 19, 5, 21, 12, 11, 8, 10, 7, 2, 14

The confidence interval is obtained using the relation :

Mean ± margin of error

Margin of Error = Zcritical * s/sqrt(n)

Using a calculator, the mean value of the data and standard deviation above is :

Mean = Σx / n ; n = sample size

Mean, xbar = 281 / 21

Mean = 13.38

Standard deviation , s = sqrt[Σ(X - xbar)²/ n-1]

Standard deviation, s = 5.82

Margin of Error = Zcritical * s/sqrt(n)

Zcritical at 95% confidence level = 1.96

Margin of Error = 1.96 * 5.82/sqrt(21) = 2.489

Confidence interval :

Mean ± margin of error

13.38 ± 2.489

Lower boundary = (13.38 - 2.489) = 10.891

Upper boundary = (13.38 + 2.489) = 15.869

(10.891; 15.869)

answer: 323 is how many she has ☃☃⛄❄

step-by-step explanation:

sas (side angle side)

aa (angle angle)

sss(side side side)

sas find two sides with an angle in between

aa is finding two angles and the third will add up to 180

sss find all sides

step-by-step explanation: