Chebyshev came up with the
limits on how much or how many of the data must lie close to the mean. In specific
for any positive k, the proportion of the data that lies within k standard
deviations of the mean is at least: <span>
1 - 1/k²
<span>In this problem the
mean is 47 yrs therefore:
(47 – 17.3) = 29.7 = (76.7 - 47) </span></span>
The value of k is
calculated using the formula:<span>
29.7 / 11 = 2.7 = k</span>
So the % of gym members
aged between 19.4 and 76.6 is: <span>
1 - 1 / (2.6)² = 0.863 = 86.3 %</span>
<span>Therefore 86.3% of the
gym members are aged between 19.4 and 76.6</span>
it will be 170.6 about 60% sure
Answer:
(C) 2/3
Step-by-step explanation:
We have 4 different books, all totaling to 15. since sci-fi plus adventure (6+4) is 10, we have 10/15. 10/15 is 2/3.
Using the binomial distribution, it is found that there is a 0.0008 = 0.08% probability that she will get a passing score.
For each question, there are only two possible outcomes, either she answers it correctly, or she does not. The probability of answering a question correctly is independent of any other question, which means that the binomial distribution is used to solve this question.
Binomial probability distribution
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 18 questions, thus
![n = 18](https://tex.z-dn.net/?f=n%20%3D%2018)
- Guess one of three correct options, thus
.
A passing score is <u>at least 13 correct</u>, thus, the probability is:
![P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%2013%29%20%3D%20P%28X%20%3D%2013%29%20%2B%20P%28X%20%3D%2014%29%20%2B%20P%28X%20%3D%2015%29%20%2B%20P%28X%20%3D%2016%29%20%2B%20P%28X%20%3D%2017%29%20%2B%20P%28X%20%3D%2018%29)
Then
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 13) = C_{18,13}.(0.3333)^{13}.(0.6667)^{5} = 0.0007](https://tex.z-dn.net/?f=P%28X%20%3D%2013%29%20%3D%20C_%7B18%2C13%7D.%280.3333%29%5E%7B13%7D.%280.6667%29%5E%7B5%7D%20%3D%200.0007)
![P(X = 14) = C_{18,14}.(0.3333)^{14}.(0.6667)^{4} = 0.0001](https://tex.z-dn.net/?f=P%28X%20%3D%2014%29%20%3D%20C_%7B18%2C14%7D.%280.3333%29%5E%7B14%7D.%280.6667%29%5E%7B4%7D%20%3D%200.0001)
The others, until P(X = 18), will be approximately 0, thus:
![P(X \geq 13) = P(X = 13) + P(X = 14) = 0.0007 + 0.0001 = 0.0008](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%2013%29%20%3D%20P%28X%20%3D%2013%29%20%2B%20P%28X%20%3D%2014%29%20%3D%200.0007%20%2B%200.0001%20%3D%200.0008)
0.0008 = 0.08% probability that she will get a passing score.
A similar problem is given at brainly.com/question/24863377