Questions: Question 10 A sample of size 95 will be drawn from a population with mean 25 and standard deviation 13. Find the probability that x̄ will be between 22 and 27. 0.0668 0.9210 0.0122 0.9080 Question 11 The average number of mosquitos in a stagnant pond is 100 per square meter with a standard deviation of 16. If 9 square meters are chosen at random for a mosquito count, find the probability that the average of those counts is more than 104.8 mosquitos per square meter. Assume that the variable is normally distributed. 0.3% 81.6% 18.4% 31.6%

Question 10
A sample of size 95 will be drawn from a population with mean 25 and standard deviation 13. Find the probability that x̄ will be between 22 and 27.
0.0668
0.9210
0.0122
0.9080

Question 11
The average number of mosquitos in a stagnant pond is 100 per square meter with a standard deviation of 16. If 9 square meters are chosen at random for a mosquito count, find the probability that the average of those counts is more than 104.8 mosquitos per square meter. Assume that the variable is normally distributed.
0.3%
81.6%
18.4%
31.6%

Transcript text: sampling error 3 pts Question 10 A sample of size 95 will be drawn from a population with mean 25 and standard deviation 13. Find the probability that $\bar{x}$ will be between 22 and 27 . 0.0668 0.9210 0.0122 0.9080 Question 11 3 pts The average number of mosquitos in a stagnant pond is 100 per square meter with a standard deviation of 16 . If 9 square meters are chosen at random for a mosquito count, find the probability that the average of those counts is more than 104.8 mosquitos per square meter. Assume that the variable is normally distributed. 0.3\% $81.6 \%$ 18.4\% $31.6 \%$

Solution

Solution Steps

Step 1: Calculate the Probability for Question 10

We are given a population with mean $ \mu = 25 $ and standard deviation $ \sigma = 13 $. A sample of size $ n = 95 $ is drawn. We need to find the probability that the sample mean $ \bar{x} $ is between 22 and 27.

First, we calculate the standard error of the mean (SEM): \[ \text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{13}{\sqrt{95}} \approx 1.3333 \]

Next, we convert the sample mean bounds to Z-scores: \[ Z_{start} = \frac{22 - \mu}{\text{SEM}} = \frac{22 - 25}{1.3333} \approx -2.2493 \] \[ Z_{end} = \frac{27 - \mu}{\text{SEM}} = \frac{27 - 25}{1.3333} \approx 1.4995 \]

Using the standard normal distribution, we find: \[ P(22 < \bar{x} < 27) = \Phi(Z_{end}) - \Phi(Z_{start}) \approx \Phi(1.4995) - \Phi(-2.2493) \approx 0.9209 \]

Step 2: Calculate the Probability for Question 11

In this case, the average number of mosquitos in a stagnant pond is $ \mu = 100 $ with a standard deviation $ \sigma = 16 $. We are sampling $ n = 9 $ square meters and need to find the probability that the average count $ \bar{x} $ is more than 104.8.

First, we calculate the standard error of the mean (SEM): \[ \text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{9}} = \frac{16}{3} \approx 5.3333 \]

Next, we convert the sample mean bound to a Z-score: \[ Z = \frac{104.8 - \mu}{\text{SEM}} = \frac{104.8 - 100}{5.3333} \approx 0.9 \]

To find the probability that $ \bar{x} $ is more than 104.8, we calculate: \[ P(\bar{x} > 104.8) = 1 - P(\bar{x} \leq 104.8) = 1 - \Phi(0.9) \approx 1 - 0.8159 \approx 0.1841 \]

Final Answer

For Question 10, the probability that $ \bar{x} $ is between 22 and 27 is approximately $ 0.9209 $.
For Question 11, the probability that $ \bar{x} $ is more than 104.8 is approximately $ 0.1841 $.

Thus, the answers are: