In a pond with a large population of turtles, the percent of three different typ

Question

In a pond with a large population of turtles, the percent of three different types of turtles are presented in the table below. A random sample of 100 turtles is taken from the pond. What is the standard deviation for the sample proportion of chicken turtles? What is the standard deviation for the sample proportion of spotted turtles? What is the standard deviation for the sample proportion of snapping turtles? What is the approximate probability that the sample proportion of chicken turtles is lower than 0.25? What is the approximate probability that the sample proportion of spotted turtles falls between 45% and 55%? What is the approximate probability that the sample proportion of snapping turtles is higher than 0.25?

Explanation / Answer

Answer:

a).

p=0.3

sd= sqrt(0.3*0.7/100) =0.0458

b).

p=0.5

sd= sqrt(0.5*0.5/100) =0.05

c).

p=0.2

sd= sqrt(0.2*0.8/100) =0.04

d).

z value for 0.25, z=(0.25-0.30)/0.0458 =-1.09

P( p<0.25) = P( z < -1.09)= 0.1379

e).

z value for 0.45, z =(0.45-0.50)/0.05 =-1

z value for 0.55, z =(0.55-0.50)/0.05 =1

p( 0.45<p<0.55) = P( -1<z<1)= P( z<1)-P( z<-1)

=0.8413-0.1587

=0.6826

f).

z value for 0.25, z =(0.25-0.20)/0.04 =1.25

P( p>0.25)= P( z>1.25)

=0.1056

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