Question 13
A population of values has a normal distribution with $\mu=243.5$ and $\sigma=67.5$. You intend to draw a random sample of size $n=113$. Please show your answers as numbers accurate to 4 decimal places.
Find the probability that a single randomly selected value is less than 247.3 .
\[
P(X< 247.3)=
\]
Find the probability that a sample of size $n=113$ is randomly selected with a mean less than 247.3 .
\[
P(\bar{x}< 247.3)=
\]
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Step 1 ：Given a population of values with a normal distribution, where the mean \(\mu\) is 243.5 and the standard deviation \(\sigma\) is 67.5.

Step 2 ：We are asked to find the probability that a single randomly selected value is less than 247.3. This can be calculated using the cumulative distribution function (CDF) for a normal distribution, which is given by the formula: \[P(X < x) = \frac{1}{2}[1 + erf(\frac{x - \mu}{\sigma\sqrt{2}})]\] where erf is the error function.

Step 3 ：Substituting the given values into the formula, we get: \[P(X < 247.3) = \frac{1}{2}[1 + erf(\frac{247.3 - 243.5}{67.5\sqrt{2}})]\]

Step 4 ：Calculating the above expression, we find that \(P(X<247.3)\) is approximately 0.5224.

Step 5 ：We are also asked to find the probability that a sample of size 113 is randomly selected with a mean less than 247.3. This can be calculated using the Central Limit Theorem, which states that the distribution of sample means approaches a normal distribution as the sample size gets larger.

Step 6 ：The mean of this distribution is the same as the mean of the population, and the standard deviation is the standard deviation of the population divided by the square root of the sample size. So, the standard deviation of the sample is \(\frac{67.5}{\sqrt{113}}\).

Step 7 ：We can then use the same formula as before to find the desired probability, but with the new mean and standard deviation. So, \[P(\bar{x} < 247.3) = \frac{1}{2}[1 + erf(\frac{247.3 - 243.5}{\frac{67.5}{\sqrt{113}}\sqrt{2}})]\]

Step 8 ：Calculating the above expression, we find that \(P(\bar{x}<247.3)\) is approximately 0.7252.

Step 9 ：So, the final answers are: \[P(X<247.3) \approx \boxed{0.5224}\] and \[P(\bar{x}<247.3) \approx \boxed{0.7252}\]