Binomials Fill in the blanks using the empirical rule Normal Distribution concep

ID: 3220529 • Letter: B

Question

Binomials

Fill in the blanks using the empirical rule Normal Distribution concept: The mean Life of a tire is 30,000 km. The standard deviation is 2,000 km.

A) 68% of all tires will have a life between ____________km and __________km.

B) 95% of all tires will have a life between ____________km and __________km.

C) What percent of the tires will have a life that exceeds 26,000 km?___________

D) If a company purchased 2,000 tires, how many tires would you expect to last more than 28,000 km? ____________

mean = 30000

standard deviation = 2000

by 68 95 99.7 rule 68% confindenc interval is (mean-sd, mean+sd)

68% confindence interval is (30000-2000,30000+2000) i.e (28000, 32000)

68% of all tires will have a life between 28000 km and 32000km.

by 68 95 99.7 rule 95% confindenc interval is (mean-2*sd, mean+2*sd)

95% confindence interval is (30000-2*2000,30000+2*2000) i.e (26000, 34000)

95% of all tires will have a life between 26000 km and 34000km.

z value for 26000 is (26000-30000)/2000 = -2, correspoding p value using z table is 0.02275

P(X<26000) = 0.02275

P(X>26000) = 1-0.02275 = 0.9772

probabity that the tires will have a life that exceeds 26,000 km = 0.9772

percent of the tires will have a life that exceeds 26,000 km = 97.72%

z value for 28000 is (28000-30000)/2000 = -1, correspoding p value using z table is 0.1587

P(X<28000) = 0.1587

P(X>28000) = 1-0.1587 = 0.8413

probabity that the tires will have a life that exceeds 28000 km = 0.8413

percent of the tires will have a life that exceeds 28,000 km = 84.13%

so no of tires that were expected to last more than 28,000 km = 2000*84.13/100 =1682.6 ~1683

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