Math Problem Statement

The results of a study showed that heterosexual​ women, during​ ovulation, were significantly better at correctly identifying the sexual orientation of a man from a photograph of his face than women who were not ovulating. Near​ ovulation, on average women correctly identified the orientation of about 67​% of the 100 men shown to them. Assume that the sample distribution for this study is unimodal and symmetric and that the samples are collected randomly. If this is the probability of correctly identifying the orientation of a man in any given​ photograph, what is the probability a woman would correctly classify 77 or more of the​ men?

Solution

This problem can be approached using the normal approximation to the binomial distribution, given that we're working with a large number of trials (100). Let's break down the steps:

1. Defining the binomial distribution parameters:

The problem involves a binomial distribution where:

  • n=100n = 100 (the number of men shown to the women),
  • p=0.67p = 0.67 (the probability of correctly identifying a man's orientation for each trial).

2. Calculating the mean and standard deviation:

For a binomial distribution:

  • The mean μ\mu is calculated as: μ=np=1000.67=67\mu = n \cdot p = 100 \cdot 0.67 = 67
  • The standard deviation σ\sigma is calculated as: σ=np(1p)=1000.670.3322.114.7\sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{100 \cdot 0.67 \cdot 0.33} \approx \sqrt{22.11} \approx 4.7

3. Standardizing the value:

To find the probability of a woman correctly classifying 77 or more men, we need to standardize this value using the normal distribution approximation.

We convert 77 to a zz-score: z=Xμσ=77674.7104.72.13z = \frac{X - \mu}{\sigma} = \frac{77 - 67}{4.7} \approx \frac{10}{4.7} \approx 2.13

4. Finding the probability:

Now, we need to find the probability that z2.13z \geq 2.13. Using standard normal distribution tables or a calculator: P(Z2.13)0.0166P(Z \geq 2.13) \approx 0.0166

Thus, the probability that a woman would correctly classify 77 or more men is approximately 0.0166, or 1.66%.


Would you like more details on any of the steps? Feel free to ask questions.

Here are some related questions to explore:

  1. How does the binomial distribution differ from the normal distribution?
  2. What are the key assumptions behind using the normal approximation for the binomial distribution?
  3. How would this problem change if the probability pp were closer to 0.5?
  4. Can the same method be used for small sample sizes? Why or why not?
  5. How is the standard deviation formula for binomial distributions derived?

Tip: When using the normal approximation for a binomial distribution, it's a good idea to check that npn \cdot p and n(1p)n \cdot (1 - p) are both greater than 5 to ensure the approximation is valid.

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Math Problem Analysis

Mathematical Concepts

Binomial Distribution
Normal Approximation
Probability
Z-Scores

Formulas

Mean μ = n * p
Standard Deviation σ = √(n * p * (1 - p))
Z-score z = (X - μ) / σ

Theorems

Normal Approximation to the Binomial Distribution
Central Limit Theorem

Suitable Grade Level

Grades 10-12