30% of the fifth grade students in a large school district read below grade level. The distribution of sample proportions of samples of 100 students from this population is normal with a mean of 0.30 and a standard deviation of 0.045. Suppose that you select a sample of 100 fifth grade students from this district and find that the proportion that reads below grade level in the sample is 0.36. What is the probability that a second sample would be selected with a proportion less than 0.36?
A. 0.8932
B. 0.8920
C. 0.9032
D. 0.9048
answer D
If you do the second time completely new after putting the first bunch back in and mixing them up, then the two experiments are independent.
Now what is probability of <.36 with mean of .3 and sd of .045?
David lane:
http://davidmlane.com/hyperstat/z_table.html
.9088
To find the probability that a second sample would be selected with a proportion less than 0.36, we need to calculate the z-score and then find the corresponding probability.
The formula to calculate the z-score is:
z = (sample proportion - population mean) / (population standard deviation / sqrt(sample size))
Given that:
Population mean (μ) = 0.30
Population standard deviation (σ) = 0.045
Sample proportion (p̂) = 0.36
Sample size (n) = 100
Calculating the z-score:
z = (0.36 - 0.30) / (0.045 / sqrt(100))
z = 0.06 / (0.045 / 10)
z = 0.06 / 0.0045
z = 13.3333
Now, we need to find the probability of a z-score less than 13.3333. However, since the distribution is normal, we need to find the area under the curve to the left of the z-score.
Using a standard normal distribution table or a calculator, we find that the probability of a z-score less than 13.3333 is approximately 1 (since it is almost at the extreme right end).
Therefore, the probability that a second sample would be selected with a proportion less than 0.36 is 1.
The correct answer is not listed in the options provided.
To solve this problem, we will use the properties of the normal distribution. The distribution of sample proportions is normal with a mean equal to the population proportion (in this case, 0.30) and a standard deviation equal to the square root of (p(1-p)/n), where p is the population proportion (0.30) and n is the sample size (100).
First, we need to calculate the z-score for the sample proportion of 0.36. The formula for the z-score is (sample proportion - population proportion) / standard deviation.
z = (0.36 - 0.30) / sqrt(0.30 * (1 - 0.30) / 100)
z = 0.06 / sqrt(0.21 / 100)
z = 0.06 / 0.045
z = 1.333
Next, we need to find the probability that a second sample would be selected with a proportion less than 0.36. Since we want to find the probability of a sample proportion less than a given value, we use the z-table or a calculator to find the area under the normal distribution curve to the left of the z-score.
Looking up the z-score of 1.333 in the z-table, we find that the probability is approximately 0.9048.
Therefore, the correct answer is D. 0.9048.