Mathematics, 21.02.2020 03:28 vanessabeausoleil

An article in the Los Angeles Times (Dec. 3, 1993) reports that 1 in 200 people carry the defective gene that causes inherited colon cancer. In a sample of 1000 individuals, what is the approximate distribution of the number who carry this gene? Use this distribution to calculate the approximate probability that a. Between 5 and 8 (inclusive) carry the gene. b. At least 8 carry the gene.

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Answer from: Ab20600

a) $P(X=5) = \frac{e^{-5} 5^5}{5!}=0.1755$

$P(X=6) = \frac{e^{-5} 5^6}{6!}=0.1462$

$P(X=7) = \frac{e^{-5} 5^7}{7!}=0.1044$

$P(X=8) = \frac{e^{-5} 5^8}{8!}=0.06528$

And adding the values we got $P(5 \leq X \leq 8) =0.4914$

b) $P(X \geq 8) = 1-P(X$

$P(X=0) = \frac{e^{-5} 5^0}{0!}=0.0067$

$P(X=1) = \frac{e^{-5} 5^1}{1!}=0.0337$

$P(X=2) = \frac{e^{-5} 5^2}{2!}=0.0842$

$P(X=3) = \frac{e^{-5} 5^3}{3!}=0.1403$

$P(X=4) = \frac{e^{-5} 5^4}{4!}=0.1755$

$P(X=5) = \frac{e^{-5} 5^5}{5!}=0.1755$

$P(X=6) = \frac{e^{-5} 5^6}{6!}=0.1462$

$P(X=7) = \frac{e^{-5} 5^7}{7!}=0.1044$

$P(X \geq 8) = 1-P(X$

Step-by-step explanation:

For this case the probability of carrying a defective gene is given by:

$p =\frac{1}{200}= 0.005$

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=1000, p=0.05)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

And we know that if for a binomial distribution the valueof n is very large and the value of p is very low then we can approximate the dsitribution with a Poisson distribution with the following parameter:

$\mu = np = 1000*0.005= 5$

And for this case then we can approximate $X \sim Poi (\mu = 5)$

Part a

For this case we want to find this probability:

$P(5 \leq X \leq 8)$

And we can use the mass function for the Poisson distribution given by:

$f(x) = \frac{e^{-\mu} \mu^x}{x!}$

And we can find the individual probabilities like this:

$P(X=5) = \frac{e^{-5} 5^5}{5!}=0.1755$

$P(X=6) = \frac{e^{-5} 5^6}{6!}=0.1462$

$P(X=7) = \frac{e^{-5} 5^7}{7!}=0.1044$

$P(X=8) = \frac{e^{-5} 5^8}{8!}=0.06528$

And adding the values we got $P(5 \leq X \leq 8) =0.4914$

Part b

For this case we want this probability:

$P(X \geq 8)$

And we can use the complement rule:

$P(X \geq 8) = 1-P(X$

$P(X=0) = \frac{e^{-5} 5^0}{0!}=0.0067$

$P(X=1) = \frac{e^{-5} 5^1}{1!}=0.0337$

$P(X=2) = \frac{e^{-5} 5^2}{2!}=0.0842$

$P(X=3) = \frac{e^{-5} 5^3}{3!}=0.1403$

$P(X=4) = \frac{e^{-5} 5^4}{4!}=0.1755$

$P(X=5) = \frac{e^{-5} 5^5}{5!}=0.1755$

$P(X=6) = \frac{e^{-5} 5^6}{6!}=0.1462$

$P(X=7) = \frac{e^{-5} 5^7}{7!}=0.1044$

$P(X \geq 8) = 1-P(X$

Answer from: Quest

the system of inequity is

$y\leq -x+4$

$3y> x$

step-by-step explanation:

from the given graph it is noticed that we have two related lines. one is solid and another is dashed.

if a line passing through two points, then the equation of line is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

the solid line passing through the points (0,4) and (4,0).

$y-4=\frac{0-4}{4-0}(x-0)$

$y=-x+4$

the related equation of solid line is $y=-x+4$ .

similarly, find the related equation of dashed line. the dashed liune passing through the points (0,0) and (3,1).

$y-0=\frac{0-1}{3-0}(x-0)$

$3y=x$

the related equation of dashed line is $3y=x$

from the given graph it is noticed that the point (0,1) lies on the shaded region, therefore the point (0,1) must be satisfied by the related equations.

check the related equations by the point (0,1).

$1=-0+4$

$1=4$

this statement is true if the sign of inequality is < instead of equality. since it is an solid line therefore the sign of inequality will be $\leq$ . the required inequality is

$y\leq -x+4$

similarly, check the second related equation.

$3(1)=0$

it is a dashed line, so this statement is true if the sign of inequality is > . the required inequality is

$3y> x$

15 points write a system of inequalities to represent the shaded portion of the graph

Answer from: Quest

not me

step-by-step explanation:

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

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