Answer:
When the sample size is [tex]n_1 = 5[/tex] ,
[tex]P( X < 11) = 0.9875[/tex]
When the sample size is [tex]n_2 = 6[/tex] ,
[tex]P( X < 11) = 0.993[/tex]
Step-by-step explanation:
From the question we are told that
The mean is [tex]\mu = 9 \ min[/tex]
The standard deviation is [tex]\sigma = 2 \ min[/tex]
The first sample size is [tex]n_1 = 5[/tex]
The second sample size is [tex]n_2 = 6[/tex]
Generally the standard error of the mean is mathematically represented as
[tex]\sigma_{x} = \frac{\sigma}{ \sqrt{n} }[/tex]
=> [tex]\sigma_{x} = \frac{2}{ \sqrt{5} }[/tex]
When the sample size is [tex]n_1 = 5[/tex] ,
Generally the standard error of the mean is mathematically represented as
[tex]\sigma_{x} = \frac{\sigma}{ \sqrt{n} }[/tex]
=> [tex]\sigma_{x} = \frac{2}{ \sqrt{5} }[/tex]
=> [tex]\sigma_{x} = 0.894[/tex]
Generally the probability that the sample average amount of time taken on each day is at most 11 min is mathematically represented as
[tex]P( X < 11) = P( \frac{X - \mu }{\sigma } < \frac{11 - 9 }{0.8944 } )[/tex]
[tex]\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )[/tex]
[tex]P( X < 11) = P( Z < 2.24 )[/tex]
From the z table the area under the normal curve to the left corresponding to 2.24 is
[tex]P( Z < 2.24 ) = 0.9875[/tex]
=> [tex]P( X < 11) = 0.9875[/tex]
When the sample size is [tex]n_2 = 6[/tex] ,
Generally the standard error of the mean is mathematically represented as
[tex]\sigma_{x} = \frac{\sigma}{ \sqrt{n} }[/tex]
=> [tex]\sigma_{x} = \frac{2}{ \sqrt{6} }[/tex]
=> [tex]\sigma_{x} = 0.816[/tex]
Generally the probability that the sample average amount of time taken on each day is at most 11 min is mathematically represented as
[tex]P( X < 11) = P( \frac{X - \mu }{\sigma } < \frac{11 - 9 }{0.816 } )[/tex]
[tex]\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )[/tex]
[tex]P( X < 11) = P( Z < 2.45 )[/tex]
From the z table the area under the normal curve to the left corresponding to 2.24 is
[tex]P( Z < 2.45 ) = 0.993[/tex]
=> [tex]P( X < 11) = 0.993[/tex]
