Here is the full question:
[tex]The \ shell \ of \ the \ land \ snail \ Limocolaria \ martensiana \ has \ two \ possible \ color[/tex][tex]forms: \ streaked \ and \ pallid. \ In \ a \ certain \ population \ of \ these \ snails, \ 60\% \ o f \ the[/tex][tex]individuals \ have \ streaked \ s hells.\ 16 \ Suppose \ that \ a \ random \ sample \ of \ 10 \ snails[/tex][tex]is \ to \ be \ chosen \ from \ this \ population. \ F ind \ the \ probabilit y \ that \ the \ percentage \ of[/tex]
[tex]streaked-shelled \ snails \ in \ t he \ sample \ will \ be[/tex]
[tex](a) 50\%. \ (b) 60\%. \ (c) 70\%.[/tex]
Answer:
(a) 0.2007
(b) 0.2510
(c) 0.2150
Step-by-step explanation:
Given that:
The sample size = 10
Sample proportion= 60% 0.6
Let X represents the no of streaked-shell snails.
[tex]X \sim Binom (n =1 0, p = 0.60)[/tex]
The Probability mass function (X) is:
[tex]P(X =x)= (^n_x) p^x(1-p)^{n-x}; x = 0,1,2,3...[/tex]
The Binomial probability with mean μ = np
= 10 * 0.6
= 6
Standard deviation σ = [tex]\sqrt{np(1-p)}[/tex]
= [tex]\sqrt{10*0.6*(1-0.6)}[/tex]
= 1.55
The probability that the percentage of the streaked-shelled snails in the sample will be:
a)
[tex]P(X = 0.5) = ^nC_x p^x (1 -p) ^{(n-x)}[/tex]
[tex]= ^{10}^C_5 * (0.6)^5(1-0.6)^{10-5}[/tex]
[tex]= \dfrac{10!}{5!(10-5)!} * (0.6)^5(1-0.6)^{10-5}[/tex]
= 0.2007
b)
[tex]P(X = 0.6) = ^nC_x p^x (1 -p) ^{(n-x)}[/tex]
[tex]= ^{10}^C_6 * (0.6)^6(1-0.6)^{10-6}[/tex]
[tex]= \dfrac{10!}{6!(10-6)!} * (0.6)^6(1-0.6)^{10-6}[/tex]
= 0.2510
c)
[tex]P(X = 0.7) = ^nC_x p^x (1 -p) ^{(n-x)}[/tex]
[tex]= ^{10}^C_7 * (0.6)^7(1-0.6)^{10-7}[/tex]
[tex]= \dfrac{10!}{7!(10-7)!} * (0.6)^7(1-0.6)^{10-7}[/tex]
= 0.2150
