Can anybody help me with this kind of problem im kinda confusing with this word problem:
At a large manufacturing company the mean age of workers is 38years, and the standard deviation is 6.2. Assume the variable is normally distributed, if a random sample of 44 workers is selected, find the probability that the mean age of the workers in the sample is between 36 and 40 years.
Using this formula: Z=x-m/s
To reframe your question, you are looking for $P(36 \leq \text{sample mean age} \leq 40)$. That simplifies to $P(\text{sample mean age}\leq 40) - P(\text{sample mean age} \leq 35)$. Note that the $36$ changed to a $35$ because ages are discrete values. Also you are given the mean age of the overall population $\mu = 38$ and the corresponding standard deviation $\sigma = 6.2$ and also the sample size of $n = 44$.
Now introduce a random variable $X$ that corresponds to the mean age in your sample size and a random variable $Z=\frac{X-\mu}{\sigma}$
Then you are looking for $P(36 \leq X \leq 40)$. Now you need to standardize this sample distribution to the normal distribution. This means the following:
$P(36 \leq X \leq 40)=P(36-\mu \leq X -\mu\leq 40-\mu)=P(\frac{36-\mu}{\sigma} \leq \frac{X -\mu}{\sigma}\leq \frac{40-\mu}{\sigma})=P(\frac{36-\mu}{\sigma} \leq Z\leq \frac{40-\mu}{\sigma})$.
Now you plug in the known values for $\mu$ and $\sigma$, rewrite it just like stated in the first paragraph, meaning
$P(\frac{36-\mu}{\sigma} \leq Z\leq \frac{40-\mu}{\sigma})= P(Z\leq \frac{40-\mu}{\sigma})-P(Z\leq \frac{36-\mu}{\sigma})$. Note that know nothing changed for the second probability, since the normal distribution is continous and not discrete. So excluding one value from the calculation would not change anything. However, one could add or subtract a $\frac{1}{2}$ in the numerator of $\frac{X-\mu}{\sigma}$ to make a continuity correction, however the results should be close to equal. You are done after you check your normal distribution table for the last two values and calculate the result.