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632. An important quality characteristic of water is the con- centration of suspended solid material. Following are 60 meas- urements on suspended solids from a cer...

Question

632. An important quality characteristic of water is the con- centration of suspended solid material. Following are 60 meas- urements on suspended solids from a certain lake. Construct a stem-: ~and-leaf diagram for these data and comment O any important features that you notice. Compute the sample mean; the sample standard deviation, and the sample median: What is the 9Sth percentile of concentration? ove 55.857.0 68.7 67.3 673 42.4 65.7 29.8 58.7 52.1 62.2 69.9 66.9 59.0 54.3 54.0 73.1 8133 5

632. An important quality characteristic of water is the con- centration of suspended solid material. Following are 60 meas- urements on suspended solids from a certain lake. Construct a stem-: ~and-leaf diagram for these data and comment O any important features that you notice. Compute the sample mean; the sample standard deviation, and the sample median: What is the 9Sth percentile of concentration? ove 55.857.0 68.7 67.3 673 42.4 65.7 29.8 58.7 52.1 62.2 69.9 66.9 59.0 54.3 54.0 73.1 8133 59.9 56.9 42.8 42.4 59.6 65.8 56.3 43.3 57.4 45.3 80.1 49.7 56.1 67.2 70.7 61.,4 64.0 64.2 72.6 72.5 46.1 53.1 51.1 73.8 42.6 77.4 77.3 39.3 76.4 59.3 54.7 57.1 59.6 66.1 31.6 614 73.1 77.3 485 89.8 50.7 52.0



Answers

Construct a normal probability plot of the suspended solids concentration data in Exercise $6-40$. Does it seem reasonable to assume that the concentration of suspended solids in water from this particular lake is normally distributed?

Thin this question. I have 29 observations, and I want to see if these observations former normal distribution. Let's take the help off history. Graham. Here is the history, um that I have drawn for all of those values and I can see that this is pretty normal. So yes, they do form a normal distribution. I already have the mean here 5.44 and sanity vision is 0.22 So my expert is 5.447 and my standard deviation over here is this value. Or I can also calculate the values by putting it in this formula, and we'll come out to be 0.22095 now, at 0.5 I want to see if the mean that I have calculated if it is less than the accepted mean today off 5.517 I'm going to use a T test, and the value from a T statistic will be found out using this formula. Expire minus mu upon s by root end, and this value turns out to be negative 1.706 This is a left tail test which means, as I am checking for less than this one to the left tail test. If this is my critical value on what is the critical value for Alfa is equal to 0.5 and degree of freedom of 28. Because DFS n minus one. So degree of freedom of 28 the critical value that I get is this minus 1.701 So my critical values one minus 1.701 And what is the critical value that I have got minus 1.706 which means that it falls in this rejection region. It is somewhere around here, and it falls in the critical region or it falls in the rejection region. Hence, I will reject than Al hypothesis. And I will say that I have enough statistical evidence to say that the mean calculate back then is less than the accepted me today. So we reject. It's not

19 the critical value for Table six, the degrees of freedom equal toe end minus one, which is 19 minus one. So is equal to 18. The chi square one minus all going toe to five, which is the chi square off all 0.975 which is equal to 8.231 And the chi square off 0.25 is equal to 31.5 to 6. So for the boundaries for the Thunder D V engine, which is in minus one over Chi Square well for over two times s, which is 19 minus 1/31 190.5 to 6 times 15, which is equal to 11.334 and end minus one over. Chi squared off one minus all for over two times as she is equal to 22.1 82 So these are the boundaries for the standard deviation. For the variance, it will be the square off these values, which is 11.33 four square and 22.218 to square, which is equal to 1 to 8 0.46 um, and for 2492.41

Hands. Clear zone you mean here? So let's let effort. That's be the uniform distribution of depth on the interval 7.5 to 20 which is given to us. So we know half of facts is equal to won over 20 minus 7.5, which is equal to 0.0 and zero. Otherwise so we could write this as zero A for the interval. 7.5 and 20 and zero for otherwise the mean distribution ISS given by negative infinity to infinity. Thanks, Prince T X. We just plug in 7.5 and 20. That's times 0.0 yes, just equal to 13.75 after we simplify and solve. So for the media to calculate variants, we first have to calculate E X Square, which is equal to the same thing. But we just plug in that square. Do you? That's this is equal to seven point 5 20 x squared points early DX, which is equal to 202.0, and that we have 13.5 and 202.0. So moving on to part being, we have to recall the definition of CD of of a continuous function which is a vex, is equal to probably that big ox smaller, equal to small huts. You have integration from negative infinity to acts of fo fly do you want and we have our given Pdf for any number between 7.5 and 20. Well, so we're here No 7.5 and ence we have planes. Zero a d u I. So we no offense is able to 0.0 a minus 0.6. This is the parentheses. So after Max can be written as zero praxis smaller than 7.5 0.0, a x minus 0.6 for the interval between 7.5 and 20 and one for excess Bigger 20 for a part, seeing we have probably that observed it is US depth is, as most 10 is denoted by probability that X is smaller or equal to 10. And using part B, you can write this as two and the probability that the observed up is between 10 and 15 is denoted by just the interval between 10 and 15 which is equal. Thio 15 minus up of 10 twitches people 2.4 for a part. Dean using part a. We know the meaning. Variants. Now the standard deviation. You calculate me good. 3.6056 Just a square root of 13. We'll continue over here, so the probability that served up is within one standard deviation of the mean value. It's giving high probability 10.6844 and 17 point 3556 We just subtract and add our standard deviation, and this is equal to 17.3 farm nines 3556 minus half prop. 10 point 64 form, which is equal to point by pre 36 And the probability that the observed death is within two standard deviations. We just add and subtract two standard deviations. We do the same thing. So yeah, probability of 6.5 creating and 20 0.9612 was equal to 20 0.9612 minus, um, six point 53 18 just equal to one


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