According to a 2009 Reader's Digest article, people throw away approximately 10% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 78 grocery shoppers to investigate their behavior. What is the probability that the sample proportion exceeds 0.18?
Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations.
the answer is 0.9903 i think
To find the probability that the sample proportion exceeds 0.18, we can use the normal distribution and the standard error formula for proportions.
The standard error formula for proportions is:
SE = √(p̂ * (1 - p̂) / n)
Where:
SE = standard error
p̂ = sample proportion
n = sample size
In this case, the sample proportion is unknown, so we will use the assumed true proportion of 10% or 0.10. The sample size is 78.
Using these values, we can calculate the standard error:
SE = √(0.10 * (1 - 0.10) / 78)
SE ≈ √(0.09 / 78)
SE ≈ √0.00115
SE ≈ 0.03393
Next, we need to calculate the z-score, which measures how many standard deviations a data point is from the mean.
z = (x - μ) / σ
Where:
z = z-score
x = sample proportion
μ = assumed true proportion (0.10)
σ = standard error (0.03393)
For this problem, we want to find the probability that the sample proportion exceeds 0.18, so x = 0.18.
z = (0.18 - 0.10) / 0.03393
z ≈ 0.08 / 0.03393
z ≈ 2.3596 (rounded to 4 decimal places)
Now, we can use a z-table or a statistical calculator to find the probability associated with the z-score of 2.3596. The probability is the area under the standard normal distribution curve to the right of the z-score.
Using a z-table or calculator, we find that the probability associated with a z-score of 2.3596 is approximately 0.0096.
Therefore, the probability that the sample proportion exceeds 0.18 is approximately 0.0096, or about 0.96%.
david lane normal distribution