Please help me with question 1.
2 answers:
You might be interested in
Answer:
The approximate distribution of the number who carry this gene is approximately normal with mean and standard deviation
Step-by-step explanation:
For each person, there are only two possible outcomes. Either they carry the defective gene that causes inherited colon cancer, or they do not. The probability of a person carrying this gene is independent from other people. So the binomial probability distribution is used to solve this question.
A sample of 2500 individuals is quite large, so we use the binomial approximation to the normal to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that ,
.
An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer.
This means that
In a sample of 2500 individuals, what is the approximate distribution of the number who carry this gene?
So
So the approximate distribution of the number who carry this gene is approximately normal with mean and standard deviation
9514 1404 393
Answer:
- relative minimum -6√3 at x = -√3
- relative maximum 6√3 at x = √3
- decreasing on x < -√3 and x > √3
- increasing on -√3 < x < √3
- see below for a graph
Step-by-step explanation:
I find it convenient to draw the graph first when looking for relative extrema.
The function can be differentiated to get ...
f'(x) = -3x^2 +9
This is zero when ...
-3x^2 +9 = 0
x^2 = 3
x = ±√3 . . . . . x-values of relative extrema
Then the extreme values are ...
f(±√3) = x(9 -x^2) = (±√3)(9 -3) = ±6√3
The lower extreme (minimum) corresponds to the lower value of x (-√3), so the extrema are ...
(x, y) = (-√3, -6√3) and (√3, 6√3)
__
Since the leading coefficient is negative and the degree is odd, the function is decreasing for values of x below the minimum and above the maximum. It is increasing for values of x between the minimum and the maximum.
decreasing: x < -√3, and √3 < x
increasing: -√3 < x < √3
