probability

Answered question

2021-12-09

1. Assume a binomial probability distribution has p = 0.60 and n = 200. a. What are the mean and standard deviation? b. Is this situation one in which binomial probabilities can be approximated by the normal probability distribution? Explain. c. What is the probability of 100 to 110 successes? d. What is the probability of 130 or more successes? e. What is the advantage of using the normal probability distribution to approximate the binomial probabilities? Use part (d) to explain the advantage

Answer & Explanation

user_27qwe

user_27qwe

Skilled2022-02-16Added 375 answers

Given:

n=200

p=0.60

a. The mean of a binomial distribution is the product of the sample size n and the probability p:

μ=np=200×0.60=120

The standard deviation of a binomial distribution is the square root of the product of the sample size n and the probabilities p and q.

σ=npq=np(1p)=200(0.60)(10.60)6.9282

b. Requirements for a normal approximation for the binomial distribution: np5 and nq5

np=200(0.60)=1205

nq=n(1p)=200(10.60)=805

Thus the requirements are satisfied.

The z-score is the value (using the continuity correction) decreased by the mean np and divided by the standard

deviation npq=np(1p).

z=xnpnp(1p)=99.5200(0.60)200(0.60)(10.60)2.96

z=xnpnp(1p)=110.5200(0.60)200(0.60)(10.60)1.37

z=xnpnp(1p)=129.5200(0.60)200(0.60)(10.60)1.37

Determine the corresponding normal probability using table 1 in the appendix:

c.

P(100x110)=P(2.96<z<1.37)=P(z<1.37)P(z<2.96)=0.08530.0015=0.0838

d.

P(X130)=P(z>1.37)=1P(z<1.37)=10.9147=0.0853

e. The approximation requires a lot less calculations than the binomial model in this case.

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