Question

18. Walleye is a common game fish. Adult walleye have a length with a mean of 44 cm and a standard deviation of 4 cm and the distribution of lengths is approximately Normal. a.What percentage of walleye are between 40 and 48 cm in length? Please show all by hand steps to earn credit. b.What length corresponds to the 25th percentile of fish lengths? Please show all by hand steps to earn credit. c. If a sample of 200 Walleye was taken, what is the probability the average length of the Walleye in this sample is less than 42? Please show all by hand steps to earn credit.

Answer 1

Given that adult walleye have a length with a mean of 44 cm and a standard deviation of 4 cm and the distribution of lengths is approximately Normal. Hence we will use properties of Normal distribution and Standard Normal distribution to answer the above questions.

Let the random variable x denote the length of adult walleye. Adult walleye have a length with a mean of 44 cm & a standard deviation of 4cm de the distribution of length is approximately Normal. ja) The a-score of 40 is Z= X-M s... = 40-44 where umean T = Standard deration = -4 =- The 2 score of 48 is . -44 P( 40 < x <48) =P(40-44 < 2 L 48-44) -PL-K<z«1) = P(z<1) - P (24-1) (9) - (-) = 0 (0) - [1-000] = 200-1 = 210.841) (From a standard normal distribution table = 1.682-1 = 0.682 in length .. Percentage of walleye between 40 & 48 cm - 100 * P(404x248) = 100 x 0.682 = 68.2

(6) To compute the 25th percentile, we use the formula X= p + ZT, & we will use the standard normal distribution table, except that we will work in the opposite direction. Previously, we started with a particular "X" & used the table to find the probability. However in this case we want to start with a 25% probability & find the value of "x" that represents it. So, we begin by going into the interior of the standard normal distribution table to find the area under the curve closest to 0.25 , & from this we can determine the corresponding Z-score. Once we have this we can use the equation = pt Zr where m= 44 & 2 =4 The exact Z value holding 25% of the values below it is - 0.67448 (as calculated from the table) x=ut Zo = 44+ (-0.67448)4 = 44 - 2.69792 = 41.302 : Length that corresponds to the 25th percentile of fish lengths = 41.302 cm

c) A sample of 200 Walleye was taken. :.n200 The sampling distribution of the sample mean is approximately normal (because the original population was approximately normal) with mean M = = 44 & standard deviation 4 = 0.28284 The a-score of 42 = 42-M 5 = Sample mean or Average length) in this case = 42 - 44 - 2 x 200 = -1200 -7.0 11067812 -7.071 (approx P(x<42) =P(Z < 42-44 ) 4 200 =P(2<-7,071) = 0 ( Using Standard normal table) .. Probat . Probability that the average length of is less than 42 = 0 the Walleye in this sample.