Approximately 8% of males experience red-green color blindness. Suppose a random sample of 200 men is...

Approximately 8% of males experience red-green color blindness. Suppose a random sample of 200 men is chosen. a. Assuming the necessary conditions are satisfied, give the mean and standard deviation of the Normal model that would be used to approximate the sampling distribution of the sample proportion of red-green color blindness. b. Sketch and clearly label the sampling distribution, based on the 68-95-99.7 Rule. c. Suppose the 200 person sample had a sample proportion of 10%. What is the probability that the sample proportion would be 10% or higher? Does that proportion seem to be an unusual result or does it meet our expectations? Briefly explain

Solutions

Expert Solution

μ= E (p̂)

= p

= 0.08.

σ = SD (p̂)

= √ [p (1 – p) / n]

= √ [(0.08) (1 – 0.08) / 200]

≈ 0.01918.

As per empirical rule

since 0.1486 not lies in +-2*sigma range so it is an unusual event

orchestra answered 1 year ago

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