We know that the mean and standard deviation of the dexterity test are 250 and 65 respectively; that is we have that:

We also know that the scores are normally distributed. With this in mind.
a.
We want the probability:
![P(240to get it, we need to find the z-score for each limit. The z-score is given by:[tex]z=(x-\mu)/(\sigma)](https://img.qammunity.org/2023/formulas/mathematics/college/yuhppxijhysqbdwjl5p1t41xn2uyc7u47x.png)
Then, we have that:
![\begin{gathered} P(240Hence we have that:[tex]P(240Now that we have our probability standarized we use probability properties and the standard normal distribution, then:[tex]\begin{gathered} P(240<strong>Therefore, the probability that a person selected at random will score between 240 and 270 is 0.1818.</strong><p></p><p>b.</p><p>In this case we are not talking about the population but from a sample. Since we know that the population follows a normal distribution we know that the mean and standar deviation for the sample are given as:</p>[tex]\begin{gathered} \bar{x}=\mu \\ \sigma_{\bar{x}}=\frac{\sigma}{\sqrt[]{n}} \end{gathered}]()
Then, in this case, we have that:
![\begin{gathered} \bar{x}=250 \\ \sigma_{\bar{x}}=\frac{65}{\sqrt[]{7}}=24.568 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wz7jkmp8m8gmpssqsmvup02ekqa9fvh8bc.png)
Now, we want to know the probability:

To find it we need to find the z-value, which is given by:

Applying it to the probability we have that:
[tex]\begin{gathered} P(240<\bar{x}<270)=P(\frac{240-250}{24.568}
Therefore, the probability that the mean score of the sample lies between 240 and 270 is 0.4503