Respuesta :
Answer:
The probability your mean age will be at least 37 is approximately 26%
Step-by-step explanation:
Let X denote the ages of all new employees hired during the last 10 years , then X is normally distributed with;
a mean of 35
a standard deviation of 10.
The sample size obtained is 10 employees. This implies that the sampling distribution of the sample mean will be normal with;
a mean of 35
a standard deviation of [tex]\sqrt{10}[/tex]
The sample mean is a statistic and thus has its own distribution. Its mean is equal to the population mean, 35 and its standard deviation is equal to [tex]\frac{sigma}{\sqrt{n} }[/tex]
where sigma is the population standard deviation, 10 and n the sample size which in this case is 10. [tex]\frac{10}{\sqrt{10} }=\sqrt{10}[/tex]
We are required to find the probability that this sample mean age will be at least 37;
P(sample mean ≥ 37)
Since we know the distribution of the sample mean we simply standardize it to obtain the z-score associated with it;
P(sample mean ≥ 37)
=[tex]P(Z\geq \frac{37-35}{\sqrt{10} })=P(Z\geq0.6325)=1-P(Z<0.6325)[/tex]
=1 - 0.7365 = 0.2635
=26%
