You may need to use the appropriate appendix table or technology to answer this question. Assume a binomial probability distribution has p = 0.60 and n = 200. (a) What are the mean and standard deviation? (Round your answers to two decimal places.) meanstandard deviation (b) Is this situation one in which binomial probabilities can be approximated by the normal probability distribution? Explain. No, because np ≥ 5 and n(1 − p) ≥ 5.No, because np < 5 and n(1 − p) < 5. Yes, because np < 5 and n(1 − p) < 5.Yes, because np ≥ 5 and n(1 − p) ≥ 5.Yes, because n ≥ 30. (c) What is the probability of 100 to 110 successes? Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.) (d) What is the probability of 130 or more successes? Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.) (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 less accurate.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 normal probability distribution to approximate the binomial probabilities increases the number of calculations.The advantage would be that using the the normal probability distribution to approximate the binomial probabilities reduces the number of calculations. How would you calculate the probability in part (d) using the binomial distribution. (Use f(x) to denote the binomial probability function.) P(x ≥ 130) = f(130) + f(131) + f(132) + f(133) + + f(200) P(x ≥ 130) = f(0) + f(1) + + f(128) + f(129) P(x ≥ 130) = f(131) + f(132) + f(133) + f(134) + + f(200) P(x ≥ 130) = 1 − f(129) − f(130) − f(131) − f(132) − − f(200) P(x ≥ 130) = f(0) + f(1) + + f(129) + f(130)

You may need to use the appropriate appendix table or technology to answer this question. Assume a binomial probability distribution has p = 0.60 and n = 200. (a) What are the mean and standard deviation? (Round your answers to two decimal places.) meanstandard deviation (b) Is this situation one in which binomial probabilities can be approximated by the normal probability distribution? Explain. No, because np ≥ 5 and n(1 − p) ≥ 5.No, because np < 5 and n(1 − p) < 5. Yes, because np < 5 and n(1 − p) < 5.Yes, because np ≥ 5 and n(1 − p) ≥ 5.Yes, because n ≥ 30. (c) What is the probability of 100 to 110 successes? Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.) (d) What is the probability of 130 or more successes? Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.) (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 less accurate.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 normal probability distribution to approximate the binomial probabilities increases the number of calculations.The advantage would be that using the the normal probability distribution to approximate the binomial probabilities reduces the number of calculations. How would you calculate the probability in part (d) using the binomial distribution. (Use f(x) to denote the binomial probability function.) P(x ≥ 130) = f(130) + f(131) + f(132) + f(133) + + f(200) P(x ≥ 130) = f(0) + f(1) + + f(128) + f(129) P(x ≥ 130) = f(131) + f(132) + f(133) + f(134) + + f(200) P(x ≥ 130) = 1 − f(129) − f(130) − f(131) − f(132) − − f(200) P(x ≥ 130) = f(0) + f(1) + + f(129) + f(130)

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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QUESTION 7

You may need to use the appropriate appendix table or technology to answer this question.

Assume a binomial probability distribution has

p = 0.60

and

n = 200.

(a)

What are the mean and standard deviation? (Round your answers to two decimal places.)

meanstandard deviation

(b)

Is this situation one in which binomial probabilities can be approximated by the normal probability distribution? Explain.

No, because np ≥ 5 and n(1 − p) ≥ 5.No, because np < 5 and n(1 − p) < 5. Yes, because np < 5 and n(1 − p) < 5.Yes, because np ≥ 5 and n(1 − p) ≥ 5.Yes, because n ≥ 30.

(c)

What is the probability of 100 to 110 successes? Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.)

(d)

What is the probability of 130 or more successes? Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.)

(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 less accurate.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 normal probability distribution to approximate the binomial probabilities increases the number of calculations.The advantage would be that using the the normal probability distribution to approximate the binomial probabilities reduces the number of calculations.

How would you calculate the probability in part (d) using the binomial distribution. (Use f(x) to denote the binomial probability function.)

P(x ≥ 130) = f(130) + f(131) + f(132) + f(133) + + f(200)

P(x ≥ 130) = f(0) + f(1) + + f(128) + f(129)

P(x ≥ 130) = f(131) + f(132) + f(133) + f(134) + + f(200)

P(x ≥ 130) = 1 − f(129) − f(130) − f(131) − f(132) − − f(200)

P(x ≥ 130) = f(0) + f(1) + + f(129) + f(130)

8.

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