a half of the students smoke?
Reiny
answered
= C(10,5)(.4^5)(.6^5)
= appr .2
40% of college students smoke ???
Are those stats from the 50's ?
Step-by-Step Bot
answered
P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)
Where:
- P(X = k) is the probability that exactly k students smoke
- n is the sample size (10 students in this case)
- k is the number of students who smoke (in this case, k = 5)
- p is the probability of a student smoking (40% or 0.4)
- (n choose k) is the number of Ways to choose k students from n students, which can be calculated using the combination formula: (n! / (k! * (n - k)!))
Now, let's substitute the values into the formula:
P(X = 5) = (10 choose 5) * 0.4^5 * (1 - 0.4)^(10 - 5)
Calculating each term:
(10 choose 5) = (10! / (5! * (10 - 5)!))
= (10! / (5! * 5!))
= (10 * 9 * 8 * 7 * 6) / (5 * 4 * 3 * 2 * 1)
= 252
0.4^5 = 0.01024
(1 - 0.4)^(10 - 5) = 0.6^5 = 0.07776
Now, substituting these back into the formula:
P(X = 5) = 252 * 0.01024 * 0.07776
P(X = 5) ≈ 0.1999
Therefore, the probability that exactly half of the students smoke is approximately 0.1999 or 19.99%.
Explain Bot
answered
The binomial probability formula is:
P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)
Where:
- P(X = k) is the probability of exactly k successes
- n is the number of trials or sample size
- p is the probability of success on a single trial
- (n choose k) is the combination formula which calculates the number of ways to choose k items from a set of n items
In this case:
- n = 10 (the sample size)
- p = 0.4 (the probability of a student smoking)
- k = 5 (since half of the students would mean 5 out of 10)
Now we can plug these values into the formula and calculate the probability:
P(X = 5) = (10 choose 5) * 0.4^5 * (1 - 0.4)^(10 - 5)
Calculating the combination (10 choose 5):
(10 choose 5) = 10! / (5! * (10-5)!) = 252
Plugging in the values:
P(X = 5) = 252 * 0.4^5 * 0.6^5
Now we can simplify:
P(X = 5) = 252 * 0.01024 * 0.07776
Finally, calculating the result:
P(X = 5) = 0.1990656
Therefore, the probability that exactly half of the students smoke is approximately 0.199 or 19.9%.
