A random sample of 10 parking meters in a resort community showed the following incomes for a day.Assume the incomes are normally distributed. Find the 95% confidence interval for the true mean. Round tothe nearest cent.$3.60 $4.50 $2.80 $6.30 $2.60 $5.20 $6.75 $4.25 $8.00 $3.00
A) ($3.39, $6.01)
B) ($2.11, $5.34)
C) ($4.81, $6.31)
D) ($1.35, $2.85)

Question

Quizplus · Accepted Answer

To find the 95% confidence interval for the true mean of normally distributed incomes, we first calculate the sample mean ($\bar{x}$) and the sample standard deviation (s). For the given data:$\bar{x} = \frac{3.60 + 4.50 + 2.80 + 6.30 + 2.60 + 5.20 + 6.75 + 4.25 + 8.00 + 3.00}{10} = \frac{47.00}{10} = 4.70$To calculate the sample standard deviation, we use the formula $s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$, where $x_i$ are the individual incomes, $\bar{x}$ is the sample mean, and $n$ is the sample size.$s \approx 1.86$The 95% confidence interval for the mean is given by $\bar{x} \pm t_{\frac{\alpha}{2}} \cdot \frac{s}{\sqrt{n}}$, where $t_{\frac{\alpha}{2}}$ is the t-score for 95% confidence with $n-1$ degrees of freedom (in this case, 9), which is approximately 2.262.Thus, the margin of error (ME) is $ME = 2.262 \cdot \frac{1.86}{\sqrt{10}} \approx 1.31$.Therefore, the 95% confidence interval is $(4.70 - 1.31, 4.70 + 1.31) = (3.39, 6.01)$.

A Random Sample of 10 Parking Meters in a Resort