A sample survey interviews an SRS of 263 college women. Suppose (as is roughly true) that 75% of all college women have been on a diet within the past 12 months.
Using the Normal approximation, what is the probability (±0.001) that 82% or more of the women in the sample have been on a diet?
To find the probability that 82% or more of the women in the sample have been on a diet, we can use the Normal approximation since the sample size is large (263) and the proportion of college women who have been on a diet within the past 12 months is roughly true at 0.75.
First, we need to calculate the mean and standard deviation for the proportion of women in the sample who have been on a diet.
Mean (μ) = Population proportion = 0.75
Standard Deviation (σ) = sqrt((p * (1 - p)) / n)
= sqrt((0.75 * (1 - 0.75)) / 263)
= sqrt(0.01875 / 263)
≈ 0.0637
Now, we want to calculate the probability that 82% or more of the women in the sample have been on a diet. To do this, we need to find the probability of the Z-score being greater than or equal to the Z-score corresponding to 82% (or 0.82).
Z = (x - μ) / σ
Z = (0.82 - 0.75) / 0.0637
≈ 1.03900
Using a Z-table or a calculator, we can find the probability:
P(Z >= 1.039) = 1 - P(Z <= 1.039)
The Z-table or calculator will give you the probability corresponding to a Z-score of 1.039 (rounded to the nearest thousandth). Subtracting this value from 1 will give you the probability of observing 82% or more of the women in the sample who have been on a diet.
Note that since we are using a Normal approximation, the answer may not be exact. However, it should give you a good estimate.
Use the binomial approximation to a normal distribution.
mean = np = 263 * .75 = ?
standard deviation = √npq = (√(263 * .75 * .25) = ?
Note: q = 1 - p
Finish the calculation.
Next, use z-scores:
z = (x - mean)/sd
x = .82
Use mean and sd calculated above.
Finally, use a z-table to determine your probability (remember the question is asking "82% or more" when looking at the table).
I hope this will help get you started.