Question

A random sample of size n = 60 is selected from a binomial distribution with population proportion p = 0.25. (a) What will be the approximate shape of the sampling distribution of p? O skewed to the right O skewed to the left O normal (b) What will be the mean and standard deviation (or standard error) of the sampling distribution of p? (Round your answers to four decimal places.) C standard deviation mean (c) Find the probability that the sample proportion p is between 0.19 and 0.44. (Round your answer to four decimal places.) You may need to use the appropriate appendix table or technology to answer this question.

Answer 1

Let p denotes the sample proportion for a random sample of size n =60. Here, Let Ø - Normal(0.25, 0.25* 3+0.25)) or Ộ - Normal(0.25, 0.0559022) Now, 60 P(0.19 <p < 0.44) = P(p < 0.44) - Pp < 0.19) i Dr P – 0.250.44 – 0.25 DP – 0.250.19 – 0.25, → P(0.19 < P < 0.44) = P6.055902 < 0.055902) – P10.055902 = 0.055902 0.440 – 0.250 0.190 – 0.250 + P(0.19 < P < 0.44) = P(Z <- 0.055902 + P(0.19 < Ô < 0.44) = P(Z < 3.398823) - P(Z < -1.073313) = P(0.19 <p < 0.44) = 0.999662 – 0.141565 = 0.858096 = 0.8581 0.055902 )

The mean of the sampling distribution of $\hat{p}$ is = 0.25

The standard deviation of the sampling distribution of $\hat{p}$ is = 0.0559