Question

A particular concentration of a chemical found in polluted water has been found to be lethal to 40% of the crayfish that are exposed to the concentration for 24 hours. 22 crayfish are placed in a tank containing this concentration of chemical in the water.
(a) What is the probability that 13 or 17 survive?
(b) What is the probability that at least 17 survive?
(c) What is the probability that at most 16 survive?
(d) What number of crayfish are expected to survive?
(e) What is the variance in the number of crayfish that are expected to survive? No actual crayfish were harmed in the making of this question.

Answer 1

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:

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