Question

10 snails is to be chosen from this population. Find the probability that the percentage of streaked-shelled snails in the sample will be (a) 50%. (b) 60%. (c) 70%.

Answer 1

Here is the full question:

$The \ shell \ of \ the \ land \ snail \ Limocolaria \ martensiana \ has \ two \ possible \ color$$forms: \ streaked \ and \ pallid. \ In \ a \ certain \ population \ of \ these \ snails, \ 60\% \ o f \ the$$individuals \ have \ streaked \ s hells.\ 16 \ Suppose \ that \ a \ random \ sample \ of \ 10 \ snails$$is \ to \ be \ chosen \ from \ this \ population. \ F ind \ the \ probabilit y \ that \ the \ percentage \ of$

$streaked-shelled \ snails \ in \ t he \ sample \ will \ be$

$(a) 50\%. \ (b) 60\%. \ (c) 70\%.$

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.

$X \sim Binom (n =1 0, p = 0.60)$

The Probability mass function (X) is:

$P(X =x)= (^n_x) p^x(1-p)^{n-x}; x = 0,1,2,3...$

The Binomial probability with mean μ = np

= 10 * 0.6

= 6

Standard deviation σ = $\sqrt{np(1-p)}$

= $\sqrt{10*0.6*(1-0.6)}$

= 1.55

The probability that the percentage of the streaked-shelled snails in the sample will be:

a)

$P(X = 0.5) = ^nC_x p^x (1 -p) ^{(n-x)}$

$= ^{10}^C_5 * (0.6)^5(1-0.6)^{10-5}$

$= \dfrac{10!}{5!(10-5)!} * (0.6)^5(1-0.6)^{10-5}$

= 0.2007

b)

$P(X = 0.6) = ^nC_x p^x (1 -p) ^{(n-x)}$

$= ^{10}^C_6 * (0.6)^6(1-0.6)^{10-6}$

$= \dfrac{10!}{6!(10-6)!} * (0.6)^6(1-0.6)^{10-6}$

= 0.2510

c)

$P(X = 0.7) = ^nC_x p^x (1 -p) ^{(n-x)}$

$= ^{10}^C_7 * (0.6)^7(1-0.6)^{10-7}$

$= \dfrac{10!}{7!(10-7)!} * (0.6)^7(1-0.6)^{10-7}$

= 0.2150