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4 years ago

7

Decrease 1280 by 67%

1 answer:

blondinia [14]4 years ago

5 0

67% of 1280 is 857.6 substrack it and it's = 422.4

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A particular concentration of a chemical found in polluted water has been found to be lethal to 40% of the crayfish that are exp

Jet001 [13]

Answer:

a) $P(X=13)=(24C13)(0.4)^{13} (1-0.4)^{24-13}=0.0608$

$P(X=17)=(24C17)(0.4)^{17} (1-0.4)^{24-17}=0.0017$

$P(X=13 U X=17) = 0.0608+0.0017=0.0624$

b) $P(X \geq 17)=0.000427$

c) $P(X \geq 16)= 1-0.000427=0.9996$

d) $E(X)= np = 22*0.4=8.8$

e) $Var(X) = np(1-p)= 22*0.4*(1-0,4)=5.28$

Step-by-step explanation:

1) Previous concepts

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".

2) Solution to the problem

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

$X \sim Binom(n=24, p=0.4)$

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:

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

3) Part a

$P(X=13)=(22C13)(0.4)^{13} (1-0.4)^{22-13}=0.0336$

$P(X=17)=(22C17)(0.4)^{17} (1-0.4)^{22-17}=0.00035$

$P(X=13 U X=17) = 0.0336+0.00035=0.0340$

4) Part b

$P(X \geq 17)=P(X=17)+P(X=18)+P(X=19)+P(X=20)+P(X=21)+P(X=22)$

$P(X=17)=(22C17)(0.4)^{17} (1-0.4)^{22-17}=0.00035$

$P(X=18)=(22C18)(0.4)^{18} (1-0.4)^{22-18}=0.0000651$

$P(X=19)=(22C19)(0.4)^{19} (1-0.4)^{22-19}=0.00000914$

$P(X=20)=(22C20)(0.4)^{20} (1-0.4)^{22-20}=0.000000914$

$P(X=21)=(22C21)(0.4)^{21} (1-0.4)^{22-21}=5.81x10^{-8}$

$P(X=22)=(22C22)(0.4)^{22} (1-0.4)^{22-22}=1.76x10^{-9}$

$P(X \geq 17)=0.000427$

5) Part c

$P(X \leq 16)=1-P(X>16)=1-P(X\geq 17) = 1-[P(X=17)+P(X=18)+P(X=19)+P(X=20)+P(X=21)+P(X=22)]$

$P(X \geq 16)= 1-0.000427=0.9996$

6) Part d

The expected value is:

$E(X)= np = 22*0.4=8.8$

7) Part e

The variance is given by:

$Var(X) = np(1-p)= 22*0.4*(1-0,4)=5.28$

3 0

3 years ago

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I mean I would say it’s 4

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