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Maksim231197 [3]
3 years ago
6

A manufacturer of radial tires for automobiles has extensive data to support the fact that the lifetime of their tires follows a

normal distribution with a mean of 42,100 miles and a standard deviation of 2,510 miles. Find the probability that a randomly selected tire will have a lifetime of between 44,500 miles and 48,000 miles. Be certain that you round your z-values to two decimal places. Round your answer to 4 decimal places.(A) 0.1685 (B) 0.8315 (C) 0.1591 (D) 0.3315 (E) None of these are correct.
Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
3 0

Answer:  (C) 0.1591

Step-by-step explanation:

Given : A manufacturer of radial tires for automobiles has extensive data to support the fact that the lifetime of their tires follows a normal distribution with

\mu=42,100\text{ miles}

\sigma=2,510\text{ miles}

Let x be the random variable that represents the lifetime of the tires .

z-score : z=\dfrac{x-\mu}{\sigma}

For x= 44,500 miles

z=\dfrac{44500-42100}{2510}\approx0.96

For x= 48,000 miles

z=\dfrac{48000-42100}{2510}\approx2.35

Using the standard normal distribution table , we have

The p-value : P(44500

P(z

Hence, the probability that a randomly selected tire will have a lifetime of between 44,500 miles and 48,000 miles =  0.1591

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As given in the question,

Given expression is equal to :

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Simplify the given expression -1 5/6 x 6 1/2 to get the value of the expression:

-1 5/6 x 6 1/2

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The complete question is:

Find the value of the given expression:

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Learn more about expression here

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