The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. 1)If a random sample of 16 cars is selected, a) what is the probability that the sample mean will be between 39 and 48 minutes? b) 95% of all sample means will fall between what two values? c) 90% of the sample means will be greater than what value? 2)What sample size is needed in order for the standard deviation of the mean to be one fifth of the population standard deviation?

Answer:

1a

  [tex]P(39 < X < 48 ) = 0.8767[/tex]

1b

    95% of all sample means will fall between [tex] 40.1  <  \mu < 49.9 [/tex]

1c

    [tex]\= x = 41. 795[/tex]

2

   [tex]n = 25[/tex]

Step-by-step explanation:

From the question we are told that

   The mean is [tex]n = 45[/tex]

   The population standard deviation is  [tex]\sigma = 10[/tex]

   The sample size is n  =  16

Generally the standard error of the mean is mathematically represented as

       [tex]\sigma_{x} = \frac{ \sigma}{\sqrt{n} }[/tex]

=>    [tex]\sigma_{x} = \frac{ 10 }{\sqrt{16 } }[/tex]

=>    [tex]\sigma_{x} = 2.5[/tex]

Generally the probability that the sample mean will be between 39 and 48 minutes is

    [tex]P(39 < X < 48 ) = P( \frac{ 39 - 45}{ 2.5} < \frac{X - \mu }{\sigma } < \frac{ 48 - 45}{ 2.5} )[/tex]

=> [tex]P(39 < X < 48 ) = P(-2.4 < Z< 1.2 )[/tex]

=> [tex]P(39 < X < 48 ) = P( Z< 1.2 ) - P(Z < -2.4)[/tex]

From the z table  the area under the normal curve to the left corresponding to  1.2  and  -2.4  is

=> [tex]P( Z< 1.2 ) = 0.88493[/tex]

and  

    [tex]P( Z< - 2.4 ) = 0.0081975[/tex]

So

   [tex]P(39 < X < 48 ) = 0.88493 -0.0081975[/tex]

=> [tex]P(39 < X < 48 ) = 0.8767[/tex]

From the question we are told the confidence level is  95% , hence the level of significance is    

      [tex]\alpha = (100 - 95 ) \%[/tex]

=>   [tex]\alpha = 0.05[/tex]

Generally from the normal distribution table the critical value  of   is  

   [tex]Z_{\frac{\alpha }{2} } =  1.96[/tex]

Generally the margin of error is mathematically represented as  

      [tex]E = Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }[/tex]

=>   [tex]E = 1.96 * 2.5 [/tex]  

=>   [tex]E =4.9 [/tex]  

Generally the  95% of all sample means will fall between

      [tex]\mu -E <  and  \mu  +E[/tex]

=>   [tex]45  -4.9\  and \ 45  + 4.9[/tex]

Generally the value which  90% of sample means is  greater than is mathematically represented

      [tex]P( \= X > \= x ) = 0.90[/tex]

=>   [tex]P( \= X > \= x ) = P( \frac{\= X - \mu }{ \sigma_x} > \frac{\= x -45 }{ 2.5} ) = 0.90[/tex]

=>  [tex]P( \= X > \= x ) = P( Z > z ) = 0.90[/tex]

Generally from the z-table  the critical  value  of  0.90  is  

      [tex]z = -1.282[/tex]

      [tex]\frac{\= x -45 }{ 2.5} = -1.282[/tex]

=>   [tex]\= x = 41. 795[/tex]

Considering question 2

 Generally we are told that the standard deviation of the mean to be one fifth of the population standard deviation, this is mathematically represented as

         [tex]s = \frac{1}{5} \sigma[/tex]

  Generally the standard deviation of the sample mean is mathematically  represented as

          [tex]s = \frac{\sigma }{ \sqrt{n} }[/tex]

=>       [tex]\frac{1}{5} \sigma = \frac{\sigma }{ \sqrt{n} }[/tex]

=>       [tex]n = 5^2[/tex]

=>       [tex]n = 25[/tex]

Using the normal distribution and the central limit theorem, it is found that:

1)

a) 0.8767 = 87.67% probability that the sample mean will be between 39 and 48 minutes.

b) 95% of all sample means will fall between 40.1 and 49.9 minutes.

c) 90% of the sample means will be greater than 41.8 minutes.

2) A sample size of 25 is needed.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

In this problem:

Item 1, question a:

The probability is the p-value of Z when X = 48 subtracted by the p-value of Z when X = 39, hence:

X = 48:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{48 - 45}{2.5}[/tex]

[tex]Z = 1.2[/tex]

[tex]Z = 1.2[/tex] has a p-value of 0.8849.

X = 39:

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{39 - 45}{2.5}[/tex]

[tex]Z = -2.4[/tex]

[tex]Z = -2.4[/tex] has a p-value of 0.0082.

0.8849 - 0.0082 = 0.8767

0.8767 = 87.67% probability that the sample mean will be between 39 and 48 minutes.

Item b:

The critical value for a 95% confidence interval is |z| = 1.96, hence:

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]-1.96 = \frac{X - 45}{2.5}[/tex]

[tex]X - 45 = -1.96(2.5)[/tex]

[tex]X = 40.1[/tex]

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]1.96 = \frac{X - 45}{2.5}[/tex]

[tex]X - 45 = 1.96(2.5)[/tex]

[tex]X = 49.9[/tex]

95% of all sample means will fall between 40.1 and 49.9 minutes.

Item c:

Greater than the 10th percentile, which is X when Z has a p-value of 0.1, so X when Z = -1.28.

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]-1.28 = \frac{X - 45}{2.5}[/tex]

[tex]X - 45 = -1.28(2.5)[/tex]

[tex]X = 41.8[/tex]

90% of the sample means will be greater than 41.8 minutes.

Item 2:

This is n for which [tex]s = \frac{10}{5} = 2[/tex], hence:

[tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

[tex]2 = \frac{10}{\sqrt{n}}[/tex]

[tex]2\sqrt{n} = 10[/tex]

Simplifying both sides by 2:

[tex]\sqrt{n} = 5[/tex]

[tex](\sqrt{n})^2 = 5^2[/tex]

[tex]n = 25[/tex]

A sample size of 25 is needed.

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