One method for estimating abundance of animals is known as line-intercept sampling. The theory of this method, when applied to Alaska wolverines, predicts that the proportion p equals.453 of attempts to locate wolverine tracks should be successful. Suppose that biologists will make 100 attempts to locate wolverine tracks in random locations in Alaska. What is the mean of the sampling distribution of p with hat on top if the proportion predicted by line-intercept sampling is correct

0.6848

Step-by-step explanation:

Mean of \hat{p} = 0.453

Answer = 0.453

Standard deviation of \hat{p} :

= \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.453(1-0.453)}{100}} = 0.0498

Answer = 0.0498

P(0.0453 - 0.05 < p < 0.0453 + 0.05)

On standardising,

= P(\frac{0.0453-0.05-0.0453}{0.0498} <Z<\frac{0.0453+0.05-0.0453}{0.0498})

= P(-1.0044 < Z < 1.0044) = 0.6848

Answer = 0.6848

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )