The tread-life of snow tire can be described by a normal model with a mean of 36

Question

The tread-life of snow tire can be described by a normal model with a mean of 36,000 miles and a standard deviation of 2500 miles.

Part 3 of 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 using the equation editor by finding the corresponding z-score to 3 decimal places! (Hint - Watch video 33 for help!)

Response feedback: Your z-score should be to 3 decimal places. Write a sentence that gives your solution in context.

I need help fixing up my answer please

Selected Answer:

Response feedback: Your z-score should be to 3 decimal places. Write a sentence that gives your solution in context.

I need help fixing up my answer please

p(x > 45000) 0.10 45000- 2500 ) = 0.10 45000- = 1.2816 2500 u-45000-2500(1.2816) = 41,796 miles

Explanation / Answer

given mean =3600

standard deviation = 2500

now 10% of snow tires S.D remains same

we know z = (X^- mean )/SD

P(Xgreatear than 45000)=0.10

45000- mean / 2500 = 0.10

mean =41,796

now finding z value we get

45000 - 41796 / 2500 = 1.281

mean of life span to be improved is 41796-36000=5796

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