Mathematics, 18.11.2020 17:30 xojade

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?

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Answer from: lorrainelopez

P(39 < X < 48 ) = 0.8767

95% of all sample means will fall between $40.1 < \mu < 49.9$

$\= x = 41. 795$

n = 25

Step-by-step explanation:

From the question we are told that

The mean is n = 45

The population standard deviation is $\sigma = 10$

The sample size is n = 16

Generally the standard error of the mean is mathematically represented as

$\sigma_{x} = \frac{ \sigma}{\sqrt{n} }$

=> $\sigma_{x} = \frac{ 10 }{\sqrt{16 } }$

=> $\sigma_{x} = 2.5$

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

$P(39 < X < 48 ) = P( \frac{ 39 - 45}{ 2.5} < \frac{X - \mu }{\sigma } < \frac{ 48 - 45}{ 2.5} )$

=> P(39 < X < 48 ) = P(-2.4 < Z< 1.2 )

=> P(39 < X < 48 ) = P( Z< 1.2 ) - P(Z < -2.4)

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

=> P( Z< 1.2 ) = 0.88493

and

P( Z< - 2.4 ) = 0.0081975

P(39 < X < 48 ) = 0.88493 -0.0081975

=> P(39 < X < 48 ) = 0.8767

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

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> E = 1.96 * 2.5

=> E =4.9

Generally the 95% of all sample means will fall between

$\mu -E < and \mu +E$

=> $45 -4.9\ and \ 45 + 4.9$

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

$P( \= X \= x ) = 0.90$

=> $P( \= X \= x ) = P( \frac{\= X - \mu }{ \sigma_x} \frac{\= x -45 }{ 2.5} ) = 0.90$

=> $P( \= X \= x ) = P( Z z ) = 0.90$

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

z = -1.282

$\frac{\= x -45 }{ 2.5} = -1.282$

=> $\= x = 41. 795$

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

$s = \frac{1}{5} \sigma$

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

$s = \frac{\sigma }{ \sqrt{n} }$

=> $\frac{1}{5} \sigma = \frac{\sigma }{ \sqrt{n} }$

=> n = 5^2

=> n = 25

Answer from: Quest

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step-by-step explanation:

independent events are ones in which the probability of one event does not affect the probability of the second event.

one student is chosen from classroom a; this does not affect the chances of choosing a student from classroom b. this makes these independent events.

when one card is chosen from a standard deck and then set aside before a second card is chosen, the probability of the second card being chosen changes. this makes them dependent events.

when one student is chosen from classroom a and then a second student is chosen from classroom a, the probability of choosing the second student changes. this makes them dependent events.

when a letter is chosen from the alphabet and then a second letter is chosen, the probability of choosing the second letter changes. this makes them dependent events.

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step-by-step explanation:

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