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Anastasy [175]
3 years ago
11

In the healthy handwashing survey conducted by Bradley Corporation, a study was found that 64% of adult Americans operate the fl

usher of toilets in public restrooms with their foot. Suppose a random sample of n=20 adult Americans is obtained, and the number x who flush public toilets with their feet is recorded.
A) Explain why this is a binomial experiment.
B) Find and interpret the probability that exactly 12 flush public toilets with their foot.
C) Find and interpret the probability that at least 16 flush public toilets with their foot.
D) Would it be unusual to find more than 17 who flush public toilets with their foot? Why?
Mathematics
1 answer:
Hitman42 [59]3 years ago
8 0

Answer:

<em>A) It fulfills the condition of binomial experiment</em>

<em>B) P (x=12) =0.1678</em>

<em>C)P ( x ≥16)= 0.1011</em>

<em>d) P (x>17) =0.0096 < 0.5</em>

Step-by-step explanation:

A. The binomial probability distribution has the following four properties

1. the outcomes of each trial maybe classified into success and failure.

2) the probability of success p remains constant for all trials.

3) the successive trials are all independent.

4) the experiment is repeated a fixed number of times ,n.

<em>These all conditions are fulfilled by the given question so it is a  binomial experiment. </em>

<em></em>

B) P (x=12) = 20C12 (0.64)^12 * (1-0.64)^8= 0.1678

C) P ( x ≥16) = P (x=16) + P (x=17) + P (x=18) + P (x=19)+ P (x=20)

where

P (x=16)= 20C16 (0.64)^16 * (1-0.64)^4= 0.0645

P (x=17)= 20C17 (0.64)^17 * (1-0.64)^3= 0.0270

P (x=18)= 20C18 (0.64)^18 * (1-0.64)^2= 0.0080

P (x=19)=20C19 (0.64)^10 * (1-0.64)^1= 0.0015

P (x=20)= 20C20 (0.64)^20 * (1-0.64)^0= 0.0001

so

P ( x ≥16)= 0.0645+ 0.0270 +0.0080+ 0.0015+0.0001= 0.1011

d) P (x>17)= P (x=17) + P (x=18) + P (x=19)+ P (x=20)

                    =0.0270 +0.0080+ 0.0015+0.0001

                         =0.0096

From this we see that the probability of  more than 17  people who flush public toilets with their foot is unusual because it is far away from the mean which is supposed to be somewhere 0.5 in a given distribution.

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