Chapter 4 Assignment Sampling Eye Color Name: Based on a study by Dr. P. Sorita

Question

Chapter 4 Assignment Sampling Eye Color Name: Based on a study by Dr. P. Sorita Soni at Indiana University, we know that the eye colors in the United States are distributed as follows: 42% brown, 35% blue, 11% green, 6% gray, and 6% hazel. A) A statistics instructor collects eye color data from her students. What is the name of this sampling method? B) Identify one factor that might make this particular sample biased and not representative of the general population of people in the United States C) If one person is randomly selected, what is the probability that this person will have brown or blue eyes? Express the probability as a decimal value D) If two people are randomly selected, what is the probability that at least one of them has brown eyes? Express the probability as a decimal value. Note: Show your work. Final answers only will not be graded.

Explanation / Answer

A) If the statsitics instructor collects the eye color data from her students, this type of sampling method is known as 'Convenience Sampling' method.

B) One factor that might make this particular sample biased is that it is 'non-probability' sampling procedure. Non-probability sample cannot be statistically tested for its goodness of fit as representative of the population.

C) P(Brown or Blue Eyes) = P(Brown) + P(Blue) - P(Brown & Blue)

But student to have Brown and blue eyes is impossible or as good as saying - having brown eyes and having blue eyes are mutually exclusive events and for mutually exclusive events P(Brown & Blue) = 0,

So P(Brown eyes or Blue eyes) = P(Brown) + P(Blue) - 0 = 0.42 + 0.35 - 0 = 0.77

D) If two people are randomly selected, then probability of at least one person having brown eyes can be calculated from showing it in the table format:

Has Brown Eyes

P(Brown Eyes) = 0.42 then P(No Brown Eyes) = 1 - 0.42 = 0.58

So, the P(at least one person has brown eyes) = P(Brown Eyes) * P(No Brown Eyes) + P(No Brown Eyes) * P(Brown Eyes) + P(Brown Eyes) * P(Brown Eyes) = 0.42*0.58 + 0.58*0.42 + 0.42*0.42 = 0.2436+0.2436+0.1764=0.6636

Probability that at least one person out of two randomly picked persons having brown eyes = 0.6636

Person1 Person2 Has Brown Eyes No Brown Eyes No Brown Eyes

Has Brown Eyes

Has Brown Eyes Has Brown Eyes

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