Math Problem Statement
the sample proportion \hat{\pi} for a random sample of 60 US adults regarding whether they have not read a book in any format in the past 12 months is approximately normally distributed with a mean of 0.23 and a standard deviation of approximately 0.0543.
Solution
The problem describes the sample proportion () from a random sample of 60 U.S. adults regarding whether they haven't read a book in the past 12 months. The given values are:
- Mean () = 0.23
- Standard deviation () = 0.0543
- Sample size
Since the sample proportion is approximately normally distributed, we can use this information to answer several questions or compute probabilities.
Would you like help with calculating probabilities, constructing confidence intervals, or testing hypotheses using this information?
Here are 5 potential questions to explore:
- What is the probability that the sample proportion is less than 0.2?
- What is the probability that the sample proportion is more than 0.25?
- How do you construct a 95% confidence interval for the true population proportion?
- How do you test if the population proportion is significantly different from 0.20?
- What is the probability that the sample proportion falls between 0.2 and 0.25?
Tip: When dealing with sample proportions, it's often useful to check if the conditions for normal approximation (np and n(1-p)) are met for a more accurate approximation.
Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Statistics
Sample Proportion
Normal Distribution
Formulas
Standard Error for Proportion: SE = sqrt( (p(1-p)) / n )
Z-score formula for proportions: Z = (\hat{\pi} - p) / SE
Theorems
Central Limit Theorem
Normal Approximation to Binomial Distribution
Suitable Grade Level
Undergraduate Statistics or Advanced High School
Related Recommendation
Sample Proportion and Hypothesis Test Assumptions for a Random Sample of 62
Probability of Less Than 23% of Adults with High Blood Pressure in a Sample of 78
Evaluating Claims with Confidence Intervals: Proportion of Adults Paying for Unused Subscriptions
Constructing 90% and 95% Confidence Intervals for Population Proportion
Calculate the Sampling Distribution of Sample Proportion for Unnecessary Medical Care