B} The average number of mosquitoes in a stagnant pond is 70 per square meter with a standard deviation of 8. If 36 square meters are chosen at random for a mosquito count, find the probability that the average of those counts is more than 70.9 mosquitoes per square meter. Assume that the variable is normally distributed. Show answer in decimal form.
I came up with the answer 54.4 .... But I am not sure how I got it and I have to show my work ?
Please someone help me
To find the probability that the average mosquito count is more than 70.9 mosquitoes per square meter, we need to use the concept of the sampling distribution of the sample mean.
First, we need to calculate the standard error of the mean (SEM) using the formula:
SEM = standard deviation / √(sample size)
In this case, the standard deviation is 8 and the sample size is 36. Therefore, the SEM is:
SEM = 8 / √(36) = 8 / 6 = 1.33
Next, we need to calculate the z-score, which represents the number of standard deviations a value is away from the mean. The formula for the z-score is:
z = (x - μ) / SEM
In this case, the value of interest is 70.9 and the mean is 70. Therefore, the z-score is:
z = (70.9 - 70) / 1.33 = 0.6767
Now we can use a standard normal distribution table or a calculator to find the probability corresponding to this z-score. The probability of getting a z-score greater than 0.6767 can be found by subtracting the area under the curve from the left side of the z-score from 1:
P(z > 0.6767) = 1 - P(z < 0.6767)
Using a standard normal distribution table, we find that P(z < 0.6767) is approximately 0.7486. Therefore:
P(z > 0.6767) = 1 - 0.7486 = 0.2514
Hence, the probability that the average mosquito count is more than 70.9 mosquitoes per square meter is approximately 0.2514, or 25.14% (rounded to two decimal places).
So, the correct answer is not 54.4 but approximately 25.14%.