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

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

think about the doctor who treated the major and the speaker in the short story "in another country." put yourself in the shoes of the doctor, and based on what you read in the story, write what you think his thoughts about the major and the speaker would be. use the following vocabulary in your response and change the form of these words as needed: detached resign withered

not me

step-by-step explanation:

i wish i did i would like to get a desert ghost ball python