Assume a binomial probability distribution has p=.60\ and\ n=200. a. what

Wotzdorfg

Wotzdorfg

Answered question

2021-09-17

Assume p=.60  and  n=200 for a binomial probability distribution.
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?

Answer & Explanation

mhalmantus

mhalmantus

Skilled2021-09-18Added 105 answers

bit a
we are suppose to do first three sub parts
Given, a binomial probability distribution has
p=0.60 and n=200
a) Mean (μ)=np
μ=2000.60
μ=120
Standard deviation (σ)=nP(1P)
σ=2000.60(10.60)
σ=6.9282
bit b
b) Here nP=2000.60
nP=120>5
n(1P)=200(10.60)
=2000.40
n(1P)=80>5
both of which are more than 5
Hence normal distribution can be a good aproximation
bit c
c) Required probability is P(100<x<110)
=P(1000.5<x<110+0.5)( continiity correction)
=P(99.5<x<110.5)
=P(99.5μσ<xμσ<110.5μσ
=P(99.51206.9282<7<110.51206.9282)
=P(2.96<7<1.37)
=P(0<7<2.96)P(0<7<1.37)( by symmetry)
=0.49850.4147
=0.0838
P(100<x<110)=0.0838

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