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irga5000 [103]
3 years ago
10

Based on a​ poll, 40​% of adults believe in reincarnation. Assume that 77 adults are randomly​ selected, and find the indicated

probability. Complete parts​ (a) through​ (d) below.
a. What is the probability that exactly 66 of the selected adults believe in​ reincarnation?
The probability that exactly 66 of the 77 adults believe in reincarnation is
nothing.
​(Round to three decimal places as​ needed.)
b. What is the probability that all of the selected adults believe in​ reincarnation?
The probability that all of the selected adults believe in reincarnation is
nothing.
​(Round to three decimal places as​ needed.)
c. What is the probability that at least 66 of the selected adults believe in​ reincarnation?
The probability that at least 66 of the selected adults believe in reincarnation is
nothing.
​(Round to three decimal places as​ needed.)
d. If 77 adults are randomly​ selected, is 66 a significantly high number who believe in​ reincarnation?
A.
No​, because the probability that 66 or more of the selected adults believe in reincarnation is less than 0.05.
B.
Yes​, because the probability that 66 or more of the selected adults believe in reincarnation is greater than 0.05.
C.
Yes​, because the probability that 66 or more of the selected adults believe in reincarnation is less than 0.05.
D.
No​, because the probability that 66 or more of the selected adults believe in reincarnation is greater than 0.05
Mathematics
1 answer:
damaskus [11]3 years ago
6 0

To solve this problem, we make use of the binomial probability equation:

P = [n! / (n – r)! r!] p^r * q^(n – r)

where,

n is the total number of adult samples = 77

r is the selected number of adults who believe in reincarnation

p is the of believing in reincarnation = 40% = 0.40

q is 1 – p = 0.60

 

a. What is the probability that exactly 66 of the selected adults believe in​ reincarnation?

So we use r = 66

 

P = [77! / (77 – 66)! 66!] 0.40^66 * 0.60^(77 – 66)

P = 1.32 x 10^-16

 

b. What is the probability that all of the selected adults believe in​ reincarnation?

So we use r = 77

 

P = [77! / (77 – 77)! 77!] 0.40^77 * 0.60^(77 – 77)

P =2.28 x 10^-31

 

c.  What is the probability that at least 66 of the selected adults believe in​ reincarnation?

So we use r = 66 to 77

 

P (r=66) = 1.32 x 10^-16

P (r=67) = [77! / (77 – 67)! 67!] 0.40^67 * 0.60^(77 – 67) = 1.44 x 10^-17

P (r=68) = [77! / (77 – 68)! 68!] 0.40^68 * 0.60^(77 – 68) = 1.42 x 10^-18

P (r=69) = [77! / (77 – 69)! 69!] 0.40^69 * 0.60^(77 – 69) = 1.23 x 10^-19

P (r=70) = [77! / (77 – 70)! 70!] 0.40^70 * 0.60^(77 – 70) = 9.38 x 10^-21

P (r=71) = [77! / (77 – 71)! 71!] 0.40^71 * 0.60^(77 – 71) = 6.17 x 10^-22

P (r=72) = [77! / (77 – 72)! 72!] 0.40^72 * 0.60^(77 – 72) = 3.43 x 10^-23

P (r=73) = [77! / (77 – 73)! 73!] 0.40^73 * 0.60^(77 – 73) = 1.56 x 10^-24

P (r=74) = [77! / (77 – 74)! 74!] 0.40^74 * 0.60^(77 – 74) = 5.64 x 10^-26

P (r=75) = [77! / (77 – 75)! 75!] 0.40^75 * 0.60^(77 – 75) = 1.50 x 10^-27

P (r=76) = [77! / (77 – 76)! 76!] 0.40^76 * 0.60^(77 – 76) = 2.64 x 10^-29

P (r=77) = 2.28 x 10^-31

 

The total is the sum of all:

P(total) = 1.48 x 10^-16

 

d. If 77 adults are randomly​ selected, is 66 a significantly high number who believe in​ reincarnation?

          C.Yes​, because the probability that 66 or more of the selected adults believe in reincarnation is less than 0.05

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