Complete Question
Answer:
The probability is Right- tailed
The mean overtime hours for the sample will be at least 9.3 hours is
[tex]P(X \ge 9.3 ) = 0.3333[/tex]
Step-by-step explanation:
From the question we are told that
The mean is [tex]\mu = 9.2 \ hours[/tex]
The standard deviation is [tex]\sigma = 1.6 \ hours[/tex]
The sample size is n = 49
The sample mean is [tex]\= x = 9.3[/tex]
Generally the standard error of the mean is mathematically represented as
[tex]\sigma_{\= x} = \frac{\sigma }{\sqrt{n} }[/tex]
=> [tex]\sigma_{\= x} = \frac{1.6 }{\sqrt{49} }[/tex]
=> [tex]\sigma_{\= x} = 0.22857[/tex]
This is a right -tailed probability because the sample mean is greater than the population mean
Generally the probability that the mean overtime hours for the sample will be at least 9.3 hours is mathematically represented as
[tex]P(X \ge 9.3 ) =1 - P(X < 9.3 )[/tex]
So
[tex]P(X < 9.3 ) = P( \frac{ X - \mu }{ \sigma_{\= x }} < \frac{ 9.3 - 9.2 }{ 0.22857} )[/tex]
Here [tex]\frac{X - \mu }{\sigma _{\= x}} = Z[/tex]
[tex]P(X < 9.3 ) = P( Z < 0.43750)[/tex]
From the z-table
[tex]P(X < 9.3 ) = P( Z < 0.43750) = 0.667[/tex]
So
[tex]P(X \ge 9.3 ) =1 - 0.667[/tex]
[tex]P(X \ge 9.3 ) = 0.3333[/tex]
