Question

At a local company, the ages of all new employees hired during the last 10 years are normally distributed. The mean age is 35 years old, with a standard deviation of 10 years.

If you were to take a sampling of 10 employees, what is the probability your mean age will be at least 37? Round to the nearest percent.

Answer 1

To find the probability that the sample mean age of 10 employees is at least 37, we can use the Central Limit Theorem and standardize the sample mean. The probability is approximately 2.28%.

Step-by-step explanation:

To solve this problem, we need to use the Central Limit Theorem, which states that the sample mean of a large enough sample size will be approximately normally distributed regardless of the shape of the population distribution.

In this case, the mean age of new employees is normally distributed with a mean of 35 and a standard deviation of 10. We want to find the probability that the sample mean age of 10 employees is at least 37.

To find this probability, we first need to standardize the sample mean using the formula z = (x - μ) / (σ / sqrt(n)), where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

Using this formula, we have z = (37 - 35) / (10 / sqrt(10)) = 2 * sqrt(10).

From a standard normal distribution table, we can find that the probability of getting a z-score less than 2 * sqrt(10) is approximately 1 - 0.0228 = 0.9772. However, we want the probability of getting a sample mean at least 37, so we subtract this probability from 1 to get 1 - 0.9772 ≈ 0.0228.

Therefore, the probability that the sample mean age of 10 employees will be at least 37 is approximately 2.28%.

Answer 2

P =26%

Step-by-step explanation:

In this problem we have the ages of all new employees hired during the last 10 years of normally distributed.

We know that the mean is
`\mu = 35` years and standard deviation is
`\sigma = 10` years

By definition we know that if we take a sample of size n of a population with normal distribution, then the sample will also have a normal distribution with a mean

`\mu_m = \mu`

And with standard deviation

`\sigma_m = (\sigma)/(√(n))`

Then the average of the sample will be

`\mu_m = 35\ years`

And the standard deviation of the sample will be

`\sigma_m =(10)/(√(10)) = 3.1622`

Now we look for the probability that the mean of the sample is greater than or equal to 37.

This is

`P({\displaystyle{\overline {x}}}\geq 37)`

To find this probability we find the Z-score

`Z = \frac{{\displaystyle{\overline{x}}} -\mu}{(\sigma)/(√(n))}`

`Z = (37 -35)/((10)/(√(10))) = 0.63`

So

`P({\displaystyle{\overline {x}}}\geq 37) = P(\frac{{\displaystyle{\overline {x}}}-\mu}{(\sigma)/(√(n))}\geq(37-35)/((10)/(√(10)))) = P(Z\geq0.63)`

We know that

`P(Z\geq0.63)=1-P(Z<0.63)`

Looking in the normal table we have:

`P(Z\geq0.63)=1-0.736\\\\P(Z\geq0.63) = 0.264`

Finally P = 26%