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.

1) what is the probability a woman would correctly classify
77 or more of the​ men?

It is estimated that the probability that, at least 77 men are correctly identified by their marital status is approximately 0.0166.

For a random variable (in this case, the identification of men by their marital status) resulting in a "success" (i.e., correctly identified) or "failure" (i.e., incorrectly identified) for finite trials, which is claimed to follow a binomial distribution, is approximated to the normal probability distribution, i.e. X ∼ Binomial (c,p) → X ∼ Normal (μ, σ²), if np ≥ 5 or n (1-p)≥5.

Where:
n = Number of Trials; and

p = probability of success.

From the information given we have 100 men, that is:

n = 100,

P = 67% = 0.67

Thus,

np = 100 * 0.67 = 67 (> 5)

And n(1-p) = 100 * (1-0.67) = 33(>5), the normal approximation is possible.

Thus,

μ = np = 100 * 0.67 = 67

σ = √(np(1-p))

= √[100*0.67*(1-0.67)]

[tex]\approx[/tex] 4.7021

Let's say X = number of men identified by their marital status correctly.

For X = 77,

Z = [X-μ]/σ

= (77-67)/4.7021

[tex]\approx[/tex] 2.13

We estimate the probability that, at least 77 men are correctly identified by their marital status, i.e.,

Probability = (X ≥ 77)

P (Z≥2.13)

= P (Z≥0) - P(0≤Z<2.13)

= 0.5 - 0.483414193316395 (area table of the normal curve)

[tex]\approx[/tex] 0.0166

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