11% of all Americans live in poverty. If 47 Americans are randomly selected, find the probability that a. Exactly 6 of them live in poverty. b. At most 6 of them live in poverty. c. At least 6 of them live in poverty. d. Between 1 and 7 (including 1 and 7) of them live in poverty.

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4) Binomial Probability (showing work is required)
You must show your work for full credit. You may take a picture of your work and upload it to the "show
work" link.
Your work should include:
1. Set up the binomial parameters for each part
2. Enter the parameters into the binomial formula for each part.
11% of all Americans live in poverty. If 47 Americans are randomly selected, find the probability that
a. Exactly 6 of them live in poverty.
b. At most 6 of them live in poverty.
c. At least 6 of them live in poverty.
d. Between 1 and 7 (including 1 and 7) of them live in poverty.
Transcribed Image Text:4) Binomial Probability (showing work is required) You must show your work for full credit. You may take a picture of your work and upload it to the "show work" link. Your work should include: 1. Set up the binomial parameters for each part 2. Enter the parameters into the binomial formula for each part. 11% of all Americans live in poverty. If 47 Americans are randomly selected, find the probability that a. Exactly 6 of them live in poverty. b. At most 6 of them live in poverty. c. At least 6 of them live in poverty. d. Between 1 and 7 (including 1 and 7) of them live in poverty.
Expert Solution
Step 1

given that n= 47 , p=0.11

Find

( a ) P( X=6 ) =?

( b ) P( X<=6 ) =?

( c) P( X>=6 ) =?

( d) P( 1<=X<=7 ) =?

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