Do Ho Omari Mason-Mozilla Firefox https//www homeworkd-336183491&questionld; 168

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Do Ho Omari Mason-Mozilla Firefox https//www homeworkd-336183491&questionld; 168dlushed fahekcld-3704081xcent om Omari Mason 2/1/16 12:17 PM Homework: Week 5 Homework 40 pts Save or D of Score: 0 ot 4 pts 5.4.38 A manufacturer claims that the life span of its tires is 50,000 miles. You work for a consumer protection agency and you are testing these tires. Assume the ife spans of the tires are | 16 of 16 (15 complete) Hw Score: 90%, 36 of 40 pts Question Help I * normally distibuted You select 100 tires at randorm and test them. The mean life span is 49,746 mles. Assume a 800 Complete parts (a) through (c) (a) Assuming the manufacturer's claim is correct, what is the probablity that the mean of the sample is 49,746 miles or less? (Round to four decimal places as needed ) In

Explanation / Answer

Mean ( u ) =49746
Standard Deviation ( sd )=800
Number ( n ) = 100
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
P(X > 49746) = (49746-49746)/800/ Sqrt ( 100 )
= 0/80= 0
= P ( Z >0) From Standard Normal Table
= 0.5                  
P(X < = 49746) = (1 - P(X > 49746)
= 1 - 0.5 = 0.5                  

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