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34​% of working mothers do not have enough money to cover their health insurance deductibles. You randomly select six working mothers and ask them whether they have enough money to cover their health insurance deductibles. The random variable represents the number of working mothers who do not have enough money to cover their health insurance deductibles. Complete parts​ (a) through​ (c) below.

User Nubela
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The sub-questions for this question are:

a) construct a binomial distribution using n=6 and p=0.34

b) graph the binomial distribution using a histogram and describe it's shape

c) what values of the random variable would you consider unusual? Explain your reasoning.

Answer:

a)

P(X=0) =0.0827

P(X=1) = 0.255

P(X=2) = 0.329

P(X=3) = 0.226

P(X=4) = 0.087

P(X=5) = 0.018

P(X=6) = 0.0015

b) graph D

c) x=5 and x=6

Explanation:

a)

Formula for binomial distribution:

nCx(p^x)(q^(n-x))

Number of sample, n = 6

probability of success, p = 0.34

probability of failure, q = 1-p = 0.66

P(X=0) = 6C0(0.34^0)(0.66^6)

= 1*1*0.0827 = 0.0827

P(X=1) = 6C1(0.34^1)(0.66^5)

= 6*0.34*0.1252 = 0.255

P(X=2) = 6C2(0.34^2)(0.66^4)

= 15*0.1156*0.1897 = 0.329

P(X=3) = 6C3(0.34^3)(0.66^3)

= 20*0.0113 = 0.226

P(X=4) = 6C4(0.34^4)(0.66^2)

= 15*0.0058 = 0.087

P(X=5) = 6C5(0.34^5)(0.66^1)

= 6*0.003 = 0.018

P(X=6) = 6C6(0.34^6)(0.66^0)

= 1*0.0015 = 0.0015

b) the shape of the graph is the graph shape. Referring to the attachment, the correct graph is D

c) the unusual values would be x=6 and x=5, because those values are too small and lower than 0.05

User Nicolas Widart
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