Right. So this problem is involving catching fish in Pyramid Lake, in Nevada and you've been given some data, The likelihood of catching zero fish in a six hour window is 40 4%. The likelihood of catching one fish is 36%. The likelihood of catching two fish is 15%. three fish was four And four or more fish was 1%. So for part A we need to convert the percentages, two probabilities. So we are going to construct a probability distribution. I like my probability distributions to be a more vertical style. So we could catch no fish, one fish, two fish, three fish or four or more. And as a probability, since there is a 44 chance of catching no fish, that is translated into 44 out of 100. So our probability would be .44 For one fish, it would be .36 for two fish could be 20.15 Three, fish would be .04, and for four or more would be .01. Now, this problem also calls for you to construct a hissed a gram. So we are going to draw are vertical and horizontal axis and the horizontal axis is going to be the number of fish Caught in a six hour period. And the vertical axis is going to be our probability and we need to number our vertical axis. So I would number it as 5, 10, 15 2025, 30, 35, 40 and 45. So the second line is 10 0.20 0.30 and 0.40. So we're ready to create our towers. So for zero fish we have to get up 2.44 said somewhere like right here, So that be zero fish for one fish, we have to get to about 36. So that's about right there. for two fish we need 15, for three fish, we need to be at four And for four or more were at one. So it's just right above there. So that would be a history graham representing the fish caught um Or the probability of fish caught in a six hour period for part B. You are trying to determine the probability that a fisherman randomly selected will catch one or more. So we can define that as X is greater than or equal to one. So that would be these four. So we're going to take the probability of catching one fish plus the probability of catching too, Plus the probability of catching three and the probability of catching four And that will be a .56 probability for part C. We want the probability of catching two or more fish. So that means we're talking two or three or four. So we would take the probability of two, Add it to the probability of three and add that to the probability of catching four fish. And we would get a .20. Now for part D. You are asked to determine the mean or the expected value. And in order to find that we're going to apply the formula sigma of X times P of X. So what I'm going to do is I'm going to start our chart all over again. So we're going to have our possible Outcomes. So we could have zero fish, one fish, two fish, three fish or four plus. But it did say to round the four or more back to just plain old four and then we have our probabilities. So we had a .44 0.36 0.15 0.4 And .01. And if we look at our formula, it's telling us we need to create a column called X times P of X, which means we're taking the X and multiplying it by its corresponding probability. So if I take one times 36 two times 1, five, And I'm going to keep that going, so I'd get .12 and .04, and when I add up that column, I will get a .82. So therefore our average or are expected value Is going to be .82. So in a six hour time period at Pyramid Lake in Nevada, we would expect less than one fish caught from the shore in that six hour period. Now we're ready to tackle part E and in part E, you are trying to find the standard deviation and to find that standard deviation, we are going to apply the formula the square root of the sum of x minus mu squared times p of x. So we're gonna go back over to our chart and we're going to add on a couple more columns, so we're going to add on x minus mu And that means we're taking the x value and subtracting the average, and we're going to get a negative .82, then we're gonna take the one minus the 10.82 and we'll get a positive 0.18 and we'll keep that going to minus the 0.82 we're going to get 1.182 point 18 and 3.18 So the next thing you're going to want to do is you're going to want to square each of those because this formula over here calls for you to square that quantity, so we're going to create a column called x minus mu quantity squared. And I'm going to bring in my graphing calculator for that. So notice I have already placed my um ex values enlist one. I have placed my P. Of X values into list too. And I want to Put my next values in list three. So I'm going to come down here and I'm going to place them in their relatively quickly. I wanted to take every value Enlist one and subtract that average. And there are the values that you have in your um X minus mu column. So now I want to square each of those. So in order to square those I'm gonna say let's take everything in column three and square them. And in doing so you can see I'm going to get some values .6742. So I'm gonna keep coming back and forth here .6742. Then I get a .03-4. Then I get A13924, A 475- four. And then a 10.11- four. And the last thing we want to do now is we want to do one more column. And this column is going to be called X minus mu squared times the corresponding p of X. So that means you're taking everything in this column and multiplying it by everything in this column. So in our graphing calculator, we called this column list one, we called this column list too. This was list three. This was list four. So technically we're taking list for and multiplying it by list too. So here's my graphing calculator. And I'm going to clear out list five and Enlist five. I want to take list too and multiply it by list for And as you can see, we get these values. So I'm going to right those values down for you. So here we get point 295856 .011664 .20886 1900 96 And .101124. And the formula calls for us to add that list. So I'm going to now add up that list And when I add up that list, I'm going to get a sum of .8076. So therefore our Standard deviation is going to be the square root of 80 76. Or we are going to have a standard deviation of 89 eight seven. So just to recap real quickly here, we have taken our data and we constructed a hist a gram and a probability distribution. We found the probability of Catching one or more fish From the shore to be 56, The probability of catching two or more fish From the shore to be .20. We found the expected value, or the Average, or mean to be .8 to fish in a six hour period, and we just found our standard deviation To be .8987.