Question

question #14
a farmer has 170 apple trees on his farm. he is estimating the average weight of the apples that he sells. if the standard deviation of the apples is 0.05 ounces, determine the margin of error of a 95% confidence interval

0.04 ounces
0.06 ounces
0.10 ounces
0.08 ounces

question #15
for a population that is distributed normally with a mean of 45 and a standard deviation of 5, use the empirical rule to determine p(x < 55).

13.5%
16%
84%
97.5%

Explanation:

Step1: Recall z - score for 95% confidence interval

For a 95% confidence interval, the z - score $z = 1.96\approx2$.

Step2: Identify the standard deviation

The standard deviation $\sigma=0.05$ ounces. Since the sample size information is not used for a rough - estimate with the z - score for 95% confidence interval in a single - sample case (assuming a large enough sample or population standard deviation known), the margin of error formula is $E = z\sigma$.

Step3: Calculate the margin of error

$E=2\times0.05 = 0.10$ ounces.

Step4: Recall the Empirical Rule for normal distribution

The Empirical Rule states that for a normal distribution: about 68% of the data lies within 1 standard deviation of the mean ($\mu\pm\sigma$), about 95% lies within 2 standard deviations ($\mu\pm2\sigma$), and about 99.7% lies within 3 standard deviations ($\mu\pm3\sigma$).

Step5: Calculate the number of standard deviations from the mean

The mean $\mu = 45$ and the value $x = 55$, and the standard deviation $\sigma=5$. The number of standard deviations $z=\frac{x-\mu}{\sigma}=\frac{55 - 45}{5}=2$.

Step6: Determine the probability

The area to the left of $\mu + 2\sigma$ in a normal distribution is $0.5+0.475 = 0.975$ or 97.5% according to the Empirical Rule (50% of the data is to the left of the mean and 47.5% is between the mean and $\mu + 2\sigma$).

Answer:

Question #14: 0.10 ounces
Question #15: 97.5%