Question

The results of a study showed that heterosexual​ women, during​ ovulation, were significantly better at correctly identifying the marriage status of a man from a photograph of his face than women who were not ovulating. Near​ ovulation, on average women correctly identified the marriage status of about 66​% 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 marriage status of a man in any given​ photograph, what is the probability a woman would correctly classify 79 or more of the​ men?

Whats the probability?

Answer 1

X ~ Bin ( n , p )

Where n = 100 , p = 0.66

Mean = n p = 100 * 0.66 = 66

Standard deviation = sqrt [ n p ( 1 - p) ]

= sqrt [ 100 * 0.66 ( 1 - 0.66) ]

= 4.7371

Using normal approximation,

P(X < x) = P(Z < ( x - mean) / SD )

With continuity correction,

P(X >= 79) = P(Z > (78.5 - 66) / 1.7371)

= P(Z > 2.64)

= 0.0041 (From Z table)