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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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Question not properly presented

Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is F(x)

0 ------ x<0

x²/25 ---- 0 ≤ x ≤ 5

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Use the cdf to obtain the following.

(a) Calculate P(X ≤ 4).

(b) Calculate P(3.5 ≤ X ≤ 4).

(c) Calculate P(X > 4.5)

(d) What is the median checkout duration, μ?

e. Obtain the density function f (x).

f. Calculate E(X).

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a. P(X ≤ 4) = 16/25

b. P(3.5 ≤ X ≤ 4) = 3.75/25

c. P(4.5 ≤ X ≤ 5) = 4.75/25

d. μ = 3.5

e. f(x) = 2x/25 for 0≤x≤2/5

f. E(x) = 16/9375

Step-by-step explanation:

a. Calculate P(X ≤ 4).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(X ≤ 4) = F(x) {0,4}

P(X ≤ 4) = x²/25 {0,4}

P(X ≤ 4) = 4²/25

P(X ≤ 4) = 16/25

b. Calculate P(3.5 ≤ X ≤ 4).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(3.5 ≤ X ≤ 4) = F(x) {3.5,4}

P(3.5 ≤ X ≤ 4) = x²/25 {3.5,4}

P(3.5 ≤ X ≤ 4) = 4²/25 - 3.5²/25

P(3.5 ≤ X ≤ 4) = 16/25 - 12.25/25

P(3.5 ≤ X ≤ 4) = 3.75/25

(c) Calculate P(X > 4.5).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(4.5 ≤ X ≤ 5) = F(x) {4.5,5}

P(4.5 ≤ X ≤ 5) = x²/25 {4.5,5}

P(4.5 ≤ X ≤ 5)) = 5²/25 - 4.5²/25

P(4.5 ≤ X ≤ 5) = 25/25 - 20.25/25

P(4.5 ≤ X ≤ 5) = 4.75/25

(d) What is the median checkout duration, μ?

Median is calculated as follows;

∫f(x) dx {-∝,μ} = ½

This implies

F(x) {-∝,μ} = ½

where F(x) = x²/25 for 0 ≤ x ≤ 5

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x²/25 {0,μ} = ½

μ² = ½ * 25

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μ = √12.5

μ = 3.5

e. Calculating density function f (x).

If F(x) = ∫f(x) dx

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f. Calculating E(X).

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E(x) = 2(2/5)³/75

E(x) = 16/9375

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