A Statistics class is estimating the mean height of all female students at their college. They collect a random sample of 36 female students and measure their heights. The mean of the sample is 65.3 inches. The standard deviation is 5.2 inches. Use the T-distribution Inverse Calculator applet to answer the following question. What is the 90% confidence interval for the mean height of all female students in their school?

a. (56.5, 74.1)
b. (63.6, 67.0)
c. (63.8, 66.8)
d. (63.9, 66.7)

d. ( 63.9, 66.7)

Therefore at 90% confidence interval (a,b) = ( 63.9, 66.7)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 65.3

Standard deviation r = 5.2

Number of samples n = 36

Confidence interval = 90%

z(at 90% confidence) = 1.645

Substituting the values we have;

65.3+/-1.645(5.2/√36)

65.3+/-1.645(0.866667)

65.3+/-1.42567

65.3+/-1.4

= ( 63.9, 66.7)

Therefore at 90% confidence interval (a,b) = ( 63.9, 66.7)

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.

hey! so 1 (25) out of 4 (100) is 25 so your answer is 25%. it's like quarters!