The Insurance Institute for Highway Safety publishes data on the total damage suffered by compact automobiles in a series of controlled, low-speed collisions. The cost for a sample of 9 cars, in hundreds of dollars, is provided below: 10,6,8,10,4,3.5,7.5,8,9 (a) What is the median cost of the total damage suffered for this sample of cars? (b) What is the first quartile for the above data? (c) What is the interquartile range of the above data? (d) What is the mean of the total damage suffered for this sample of cars? (e) Using the correct units, what is the value of the variance?

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Step-by-step explanation:

Given the cost data: 10,6,8,10,4,3.5,7.5,8,9

A.) Median cost :

Sorted data:

3.5, 4, 6, 7.5, 8, 8, 9, 10, 10

0.5(n + 1)th term ; n = number of samples

0.5(9 + 1)th term

= 0.5(10)th term = 5th term

Median = 5th term = 8

B.) First quartile (Q1) :

0.25(n + 1)th term ; n = number of samples

0.25(9 + 1)th term

= 0.25(10)th term = 2.5th term

Q1 = (2nd term + 3rd term) / 2 = (4+6)/2 = 10/2 = 5

Q1 = 5

C.) interquartile range = (Q3 - Q1)

Q3 = 0.75(n + 1)th term ; n = number of samples

0.75(9 + 1)th term

= 0.75(10)th term = 7.5th term

Q3 = (7th term + 8th term) / 2 = (9+10)/2 = 19/2 = 9.5

Q3 = 9.5

Q3 - Q1 = (9.5 - 5) = 4.5

D.) Mean (μ) :

μ = ΣX / n

n = sample size

μ =(10+ 6 + 8 + 10+ 4 + 3.5 + 7.5 + 8 + 9) / 9

μ = 66 / 9

μ = 7.33

3.) Population variance (s²)

Σ(x - μ)² / n - 1

=[(10 - 7.33)^2 + (6 - 7.33)^2 + (8 - 7.33)^2 + (10 - 7.33)^2 + (4 - 7.33)^2 + (3.5 - 7.33)^2 + (7.5 - 7.33)^2 + (8 - 7.33)^2 + (9 - 7.33)^2] / 9

= 45.5001 / 8

= 5.687

cosine

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