Question

In a certain city, it is known that 16.98% of the residents have red hair Suppose we select a random sample of 716 members. What is the probability that ever 15 829% of the members in this sample have red hair? Select one a. 0.7939 b. 0.2061 c08200 Od 0.0000 We cannot answer this question with the information given

Answer 1

Sample size = n = 716

p = Proportion of residents having red hair = 16.98% = 0.1698

Since, n* p > 10 and n*(1-p) > 10, we can assume the sampling distribution of proportion to be approximately normal with

Mean = p = 0.1698

Standard deviation = $\sqrt{\frac{p*(1-p)}{n}}=\sqrt{\frac{0.1698*(1-0.1698)}{716}}=0.014$

So,

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P(p > 0.15829) = P(Z > 0.15829 – 0.1698 0.014

Converting normal variate p into standard normal variate z,

  

It can be obtained using the z table by finding the area corresponding to the z = -0.8 at 0.02 probability.

= 0.2061

Hence, option (b) is the correct option.