DMy Hankeye Homexlopic Ch icussionGprobability equation xCSign in or Sign Up Che

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DMy Hankeye Homexlopic Ch icussionGprobability equation xCSign in or Sign Up Che X s//canvas edu/courses/15301/discussion topics/55140 MAT-156-15>Discussions Ch 5 Discussion ts This is a graded discussion: 5 points possible Ch 5 Discussion If 60% of all women are employed outside the home, find the probability that in a sample of 10 women, a.Exactly 7 are employed outside the home bAt least 8 are employed outside the home c At most 3 are not employed outside the home due Oct 13 at 11:30pm Scarch entries or author Unread D 6 Reply Replies are only visible to those who have posted at least one reply Previous Next+ o search 0 ID

Explanation / Answer

Let random variable X is the event that women are employed outside the home.

P(X) = 0.60

number of samples ,n = 10 which is fixed.

We use binomial distribution to find probabilities,

a)

P(X = 7 ) = nCx * px (1 - p)n - x

= 10C7 * (0.60)7*(1 - 0.60)10 - 7

= 0.2150

P(X = 7 ) = 0.2150

b)

P(At least 8 are employed outside the home) = P(x > = 8)

P( x > = 8) = P(x = 8) * P( x =9 ) * P(x =10)

Similarly to above,

P(x =8) = 0.1209

P(x = 9) = 0.0403

P(x = 10) = 0.0060

P( x > = 8) = P(x = 8) * P( x =9 ) * P(x =10)

= 0.1209 * 0.0403 * 0.0060

= 0.0000298

P(At least 8 are employed outside the home) = P( x > = 8)   = 0.0000298

c)

At most 3 are not employed outside home = 1 -  At most 3 are employed outside home

P( At most 3 are employed outside home) = P(x =0 ) + P(x =1) + P(x = 2) + P(x =3)

= 0.0001 + 0.0016 + 0.0106 + 0.0425

= 0.0548

P( At most 3 are employed outside home) = 0.0548

P(At most 3 are not employed outside home) = 1 - P(At most 3 are employed outside home)

= 1 - 0.0548

= 0.9452

P(At most 3 are not employed outside home) = 0.9452

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