Question

sals are out of date. Please sign in as re e du so we can verify your subscription An insurance company crashed four cars in succession at 5 miles per hour The cost of repair for each of the four crashes was $416, $459, $417, $216 Compute the range, sample variance, and sample standard deviation cost of repair. The range is $ sample variance and sample standard deviation cost of repair

Answer 1

Solution:

Given: The cost of repairs for each of the four crashes : $416 , $459 , $417 , $216

Part a)

Range=L - S

Where L = Largest observation = 459 and S = Smallest Observation = 216

thus

Range = $459 - $216

Range = $ 243

Part b)

Sample variance:

2 Σα - (Σα): /n η – 1

thus we need to make following table:

x	x^2
416	173056
459	210681
417	173889
216	46656
Στ = 1508	Σ ? = 604282

Thus

2 Σα - (Σα): /n η – 1

| 604282-(1508524 4-1

604282 - 568516 S2 =

35766

$2 = 11922

Part c)

Sample standard Deviation : s

s= V s2

s= V11922

s = 109.1879