The average number of miles (in thousands) that a car's tire will function before needing replacement is 64 and the standard deviation is 20. Suppose that 49 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution.What is the distribution of XX? XX ~ N(,)What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 65.6 and 67.8. For the 49 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 65.6 and 67.8. For part d), is the assumption that the distribution is normal necessary? YesNo

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Asked Nov 14, 2019
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The average number of miles (in thousands) that a car's tire will function before needing replacement is 64 and the standard deviation is 20. Suppose that 49 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution.

  1. What is the distribution of XX? XX ~ N(,)
  2. What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)
  3. If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 65.6 and 67.8. 
  4. For the 49 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 65.6 and 67.8. 
  5. For part d), is the assumption that the distribution is normal necessary? YesNo
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Expert Answer

Step 1

Hey, since there are multiple subparts posted, we will answer first four question. If you want any specific question to be answered then please submit that question only or specify the question number in your message.

(a)

The distribution X follows normal with mean µ and standard deviation σ. That is, X~N (µ, σ).

Here, the value of mean is 64 and the standard deviation is 20.

Hence, the distribution X follows normal with mean 64 (µ) and standard deviation 20 (σ). That is, X~N (64,20).

(b)

The distribution X-bar follows normal with mean µX-bar and standard deviation σX-bar. That is, X-bar~N (µX-bar, σX-bar).

The value of mean and standard deviation is,

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= 64 σ 20 V49 = 2.8571

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Step 2

Thus, the distribution X-bar follows normal with mean 64 (µX-bar) and standard deviation 2.8571 (σX-bar). That is, X-bar~N (60,2.8571).

(c)

The probability that the number of miles (in thousands) before it will need replacement is between 65.6 and 67.8is,

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(X-ZX- P(65.6< X<67.8) = P 65.6-64 = P 67.8-64 <Z < 20 20 =P(0.08< Z <0.19) =P(Z<0.19)- P(<0.08)[ Use Standard Normal Table = 0.57535-0.53188 =0.04347 0.0435

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Step 3

Thus, the probability that the number of miles (in thousands) before it will need replacement is between 65.6 and 67.8 is 0.0435.

(d)

The p...

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X - P(65.6<X <67.8)= P X -u <Z< 65.6 64 =P 67.8-64 <Z< 2.8571 2.8571 =P(0.56 <Z <1.33) =P(Z<1.33)-P(Z < 0.56)[ Use Standard Normal Table] =0.90824- 0.71226 0.19598 0.1960

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