In a sample of 500 college students fifty percent of all college students attend schools within 50 miles of
their homes.
The probability that the population proportion will be between 0.45 and 0.55 is
(1) 0.4875
(2) 0.9818
(3) 0.4909
(4) 0.9750
(5) 0.5000
0.4909
To find the probability that the population proportion will be between 0.45 and 0.55, we can use the normal distribution.
Let's define the following variables:
- p: proportion of college students attending schools within 50 miles of their homes
- n: sample size (500, in this case)
- σ: standard deviation of the population proportion (which can be approximated by a formula: σ = sqrt(p*(1-p)/n))
- α: level of significance (which is typically 0.05 for a 95% confidence level)
First, let's find the standard deviation (σ) using the given information:
p = 0.50 (50% of college students)
n = 500 (sample size)
σ = sqrt(0.5*(1-0.5)/500) = sqrt(0.25/500) = sqrt(0.0005) = 0.0224 (approximately)
Next, we need to calculate the z-scores for the lower and upper limits of the range (0.45 and 0.55):
z1 = (0.45 - p) / σ = (0.45 - 0.50) / 0.0224 = -0.05 / 0.0224 = -2.23 (approximately)
z2 = (0.55 - p) / σ = (0.55 - 0.50) / 0.0224 = 0.05 / 0.0224 = 2.23 (approximately)
Now, we can use a z-table or a calculator to find the area under the normal curve between these z-scores. The area represents the probability.
Looking up the z-scores in a standard normal distribution table, we find that the area to the left of -2.23 is approximately 0.0125, and the area to the left of 2.23 is approximately 0.9875.
To find the probability between the two z-scores, we subtract the area to the left of the lower z-score from the area to the left of the upper z-score:
probability = 0.9875 - 0.0125 = 0.9750
Therefore, the answer is option (4) 0.9750.