(B) 16.5π
<u>Explanation:</u>
The shape given is a cylinder
The height, h of the cylinder = 4cm
Diameter of the cylinder = 3cm
Radius, r = 1.5 cm
Surface area of cylinder, A = ?
We know:
Surface area of cylinder = 2πrh + 2πr²
Therefore, the surface area would be 16.5π
Answer:
The correct 95% confidence interval is (8.4, 8.8).
Step-by-step explanation:
The information provided is:
(a)
The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:
The 95% confidence interval for the average satisfaction score is computed as:
8.6 ± 1.96 (2.2)
This confidence interval is incorrect.
Because the critical value is multiplied directly by the standard deviation.
The correct interval is:
(b)
The (1 - <em>α</em>)% confidence interval for the parameter implies that there is (1 - <em>α</em>)% confidence or certainty that the true parameter value is contained in the interval.
The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true there is a 95% confidence that the true parameter value is contained in this interval.
The mistake is that the student concluded that the sample mean is contained in between the interval. This is incorrect because the population is predicted to be contained in the interval.
(c)
The (1 - <em>α</em>)% confidence interval for population parameter implies that there is a (1 - <em>α</em>) probability that the true value of the parameter is included in the interval.
The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true mean satisfaction score is contained between 8.4 and 8.8 with probability 0.95 or 95%.
Thus, the students is not making any misinterpretation.
(d)
According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.
In this case the sample size is,
<em>n </em>= 500 > 30
Thus, a Normal distribution can be applied to approximate the distribution of the alumni ratings.