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(1) The scores on an aptitude test for finger dexterity are normally distributed with mean 250 andstandard deviation 65.a What is the probability that a person selected at random will score between 240 and 270 on thetest?b) Test is given to a random sample of seven people. What is the probability that the mean score,I for the sample will be between 240 and 270?

(1) The scores on an aptitude test for finger dexterity are normally distributed with-example-1

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


\begin{gathered} \mu=250 \\ \sigma=65 \end{gathered}

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)

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}

Now, we want to know the probability:


P(240<\bar{x}<270)

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


z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}

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
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