Final Answer:
Option (a) is correct because a sample size of 50 is sufficient to estimate the population mean within 5 of its actual value with 95% confidence, based on the given conditions and calculations.
Step-by-step explanation:
In order to calculate the 95% confidence limits for the population mean, we use the formula for the confidence interval:
![\[ \bar{x} \pm Z \left((s)/(√(n))\right) \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/15n2pfiduafbx32wtwwo3epz6kmqzlxhol.png)
Where:
- s is the standard deviation of the sample,
- n is the sample size, and
- Z is the Z-score associated with the desired confidence level.
Given in the question,
, and
. The Z-score for a 95% confidence interval is approximately 1.96.
![\[ 145 \pm 1.96 \left((40)/(√(60))\right) \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/pbfd45g58ups7hfrv50mwnxtlb9ouvrrqq.png)
Calculating this, we get the confidence interval for the population mean.
Now, to find the required sample size to estimate the population mean within 5 of its actual value with 95% confidence, we use the formula:
![\[ n = \left((Z * s)/(E)\right)^2 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/elu57a27y3bo9a9f3isxgrlpf0372v9c5c.png)
Where:
- E is the margin of error (5 in this case).
Substituting the given values, we get:
![\[ n = \left((1.96 * 40)/(5)\right)^2 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/2h2w0g4wpn0mqjj5h618dczkxalr7qagp8.png)
Calculating this, we find that the required sample size is approximately 50. Therefore, the correct answer is option (a) 50.