An online magazine finds that the mean number of typographical errors per page is five. Find the probability that the number of typographical errors found on any given page is (a) exactly five, (b) less than five, and (c) exactly zero.
Table of contents
- 1. Intro to Stats and Collecting Data55m
- 2. Describing Data with Tables and Graphs1h 55m
- 3. Describing Data Numerically1h 45m
- 4. Probability2h 16m
- 5. Binomial Distribution & Discrete Random Variables2h 33m
- 6. Normal Distribution and Continuous Random Variables1h 38m
- 7. Sampling Distributions & Confidence Intervals: Mean1h 53m
- 8. Sampling Distributions & Confidence Intervals: Proportion1h 12m
- 9. Hypothesis Testing for One Sample2h 19m
- 10. Hypothesis Testing for Two Samples3h 22m
- 11. Correlation1h 6m
- 12. Regression1h 4m
- 13. Chi-Square Tests & Goodness of Fit1h 20m
- 14. ANOVA1h 0m
5. Binomial Distribution & Discrete Random Variables
Binomial Distribution
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A biologist is monitoring a large bird sanctuary where a particular bird species is known to have a 70% success rate for each nesting attempt (at least one chick fledges from the nest). This season, she observes 500 independent nesting attempts across the sanctuary.
(B) What is the probability that less than 330 attempts are successful?
A
B
C
D

1
Step 1: Identify the type of distribution. Since the problem involves a fixed number of independent trials (500 nesting attempts) with a constant probability of success (70%), this is a binomial distribution. The parameters are n = 500 (number of trials) and p = 0.7 (probability of success).
Step 2: Define the random variable. Let X represent the number of successful nesting attempts. X follows a binomial distribution, X ~ Binomial(n=500, p=0.7). The goal is to find P(X < 330), the probability that fewer than 330 attempts are successful.
Step 3: Approximate the binomial distribution using a normal distribution. For large n, the binomial distribution can be approximated by a normal distribution with mean μ = n * p and standard deviation σ = sqrt(n * p * (1 - p)). Calculate μ = 500 * 0.7 and σ = sqrt(500 * 0.7 * 0.3).
Step 4: Apply the continuity correction. Since the binomial distribution is discrete and the normal distribution is continuous, apply a continuity correction by finding P(X < 330) as P(X ≤ 329.5) in the normal distribution.
Step 5: Standardize the value for the normal distribution. Use the z-score formula z = (X - μ) / σ to convert X = 329.5 into a z-score. Then, use standard normal distribution tables or software to find the cumulative probability corresponding to this z-score, which gives P(X < 330).
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