Question

Assume a binomial probability distribution has p = 0.70 and n = 300. (a) What are the mean and standard deviation? (Round your answers to two decimal places.) mean Incorrect: Your answer is incorrect. standard deviation Incorrect: Your answer is incorrect. (b) Is this situation one in which binomial probabilities can be approximated by the normal probability distribution? Explain. Yes, because np ≥ 5 and n(1 − p) ≥ 5. Yes, because n ≥ 30. No, because np < 5 and n(1 − p) < 5. Yes, because np < 5 and n(1 − p) < 5. No, because np ≥ 5 and n(1 − p) ≥ 5. Correct: Your answer is correct. (c) What is the probability of 190 to 200 successes? Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.) Incorrect: Your answer is incorrect. (d) What is the probability of 220 or more successes? Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.) Incorrect: Your answer is incorrect. (e) What is the advantage of using the normal probability distribution to approximate the binomial probabilities? The advantage would be that using the normal probability distribution to approximate the binomial probabilities makes the calculations more accurate. The advantage would be that using the the normal probability distribution to approximate the binomial probabilities reduces the number of calculations. The advantage would be that using the normal probability distribution to approximate the binomial probabilities makes the calculations less accurate. The advantage would be that using the normal probability distribution to approximate the binomial probabilities increases the number of calculations. Correct: Your answer is correct. How would you calculate the probability in part (d) using the binomial distribution. (Use f(x) to denote the binomial probability function.) P(x ≥ 220) = f(0) + f(1) + + f(219) + f(220) P(x ≥ 220) = f(220) + f(221) + f(222) + f(223) + + f(300) P(x ≥ 220) = f(221) + f(222) + f(223) + f(224) + + f(300) P(x ≥ 220) = 1 − f(219) − f(220) − f(221) − f(222) − − f(300) P(x ≥ 220) = f(0) + f(1) + + f(218) + f(219) Correct: Your answer is correct.

Answer 1

Answer:

This is a binomial distribution question with

n = 300

p = 0.70

q = 1 - p = 0.30000000000000004

This binomial distribution can be approximated as Normal distribution since

np > 5 and nq > 5

Since we know that

a)

b)

5) Yes because np > 5 and n(1-p) > 5

c)

This implies that

P(190.0 < x < 200.0) = P(-2.5197 < z < -1.2599) = 0.098

d) x = 220

P(x > 220.0)=?

The z-score at x = 220.0 is,

z = 1.2599

This implies that

P(x > 220.0) = P(z > 1.2599) = 1 - 0.8961472806260209

P(x > 220.0) = {0.104}

e)

The advantage would be that normal distribution to approximate the binomial probability reduces the number of calculation