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

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 =
`√(np(1-p))`

=
`√(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)`

`= (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)`

`= (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)`

`= (10!)/(7!(10-7)!) * (0.6)^7(1-0.6)^(10-7)`

= 0.2150