The tread life of a snow tire can be described by a normal model with a mean of

Question

The tread life of a snow tire can be described by a normal model with a mean of 36,000 miles and a standard deviation of 2500 miles
Part 1 How probable is it for our tires to last 34,000 miles or less? Explain using statistical reasoning. Show all calculations, this should include a z-score and the probability.
Part 2 Approximately _________ of tires (to the nearest tenth of a percent) will last between 30,000 miles and 40,000 miles.
Part 3 The snow tire company wants to advertise that 10% of their snow tires last longer than 45,000 miles. If the standard deviation remains at 2500 miles, what must the mean life span of the tread be improved to? Show work by finding the corresponding z-score to 3 decimal places. The tread life of a snow tire can be described by a normal model with a mean of 36,000 miles and a standard deviation of 2500 miles
Part 1 How probable is it for our tires to last 34,000 miles or less? Explain using statistical reasoning. Show all calculations, this should include a z-score and the probability.
Part 2 Approximately _________ of tires (to the nearest tenth of a percent) will last between 30,000 miles and 40,000 miles.
Part 3 The snow tire company wants to advertise that 10% of their snow tires last longer than 45,000 miles. If the standard deviation remains at 2500 miles, what must the mean life span of the tread be improved to? Show work by finding the corresponding z-score to 3 decimal places.
Part 1 How probable is it for our tires to last 34,000 miles or less? Explain using statistical reasoning. Show all calculations, this should include a z-score and the probability.
Part 2 Approximately _________ of tires (to the nearest tenth of a percent) will last between 30,000 miles and 40,000 miles.
Part 3 The snow tire company wants to advertise that 10% of their snow tires last longer than 45,000 miles. If the standard deviation remains at 2500 miles, what must the mean life span of the tread be improved to? Show work by finding the corresponding z-score to 3 decimal places.

Explanation / Answer

given mean = 36000 standard devaition s = 2500

we know that standard z formulea z = [ x- u ] / standard devaition

1) p(x< =34000)

z= ( 34000-36000) / 2500 =- 2000/2500 = -0.8

p(z<-0.8) = area is left of -0.8

p(x< =34000) = p(z<-0.8) =0.2119 = 21.19%

2 ) p(30000<x<40000)

we know z = [x-u]/s

z1= [30000-36000]/2500=-2.40

z2 =[40000-36000]/2500= 1.60

p(-2.40<z<1.60)= area in between -2.40 and 1.60

=p(z<1.60)-p(z<-2.40) =0.452-0.0082= 0.9370 = 93.7 %

so appeoximately 94 of tires

3) need to find u for p(x>45000) = 0.1

given normal distribution = 2500

p(x>xo) = 0.1 = p(x<xo)=0.9

p[(x-u)/s <( x0-u)/s ] =0.9

p(z<( x0-u)/s =0.9

( x0-u)/s = inv norm 0.9

use z table to get inv norm 0.9 =1.282

u=45000-2500*1.282=41795

u=41795

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