Question

Question 1 5 pts The weights of creatures on "Planet Zaxorn 245" are very strange. The mean weight of creatures is =3,916 pounds with a standard deviation of o = 696 pounds The weights of these creatures are normally distributed. Determine the weight (i.e. the x-value) a Zaxornian must be lighter than to be in the bottom 2% of weights. Round your answer to two decimal places. Note on "Backward Normal Table" problems: When finding a zed-value associated to a given area, use the zed value (listed to two-decimal places) that is closest to the given area. For example, if I were looking for the 68th percentile in the Normal Table, the area in the interior of the table that is closest to this area is.6808. Hence, I wold use the zed-value of z = .47 (which is connected to that given area).

Answer 1

Solution - Mean = M = 3916 Standard deviation=o=696 value of a-score that are under the normal curve of the left of it equal to 2% is given by = 2=-2005 2 =X-u -2.05 = X. 3916 696 2= 696*(-2-05] +3916 x = 2489.2 Hence, the value of wight on Zastornian must be light than to be in the bottom 1 to of wieghts = 248902 pounds. Scanned with CamScanner

under Solution & Mean=l=6214 Standard devicetwin = 0 -623 value of z-score such that area the normal curve to the right of it eqreal to it, is given by 2 = 2.33 z =d-M 2.33 =2-6214 623 [x=76650597 Hence, value of wieght a zaxornian must exceed to be in top 1% of weights - 7665.59 pounds. Scanned .with CamScanner

Solution 3 Meana Ml: 6500 - Standard deviation=o=100 Probability of observing a wight between 6312 pounds and 6688 pounds = P(6312 2446688) 2-score for (x = 6312) 2-4-4 2 = 6312 – 6500 100 22-1-88 z-score for (x=6688) 2 = 6688-6500 Too 2 -1.88 ter me P(6312 < x26688) - P(-1.884 231.83). =PL2=1.33 ) - P(22-183) From normal distribution table; P(25 1:88)- ...Plzs:-1-88) = 0.9699-0.0301 = 0.9398 Hence, propability of observing a weight between 6312 pounds and 6688 founds. L CS Scanned witF 09398 T CamScanner

Solution 4 Meana M- 7500 Standard devicilion - 0 100 probability of observing a weight belonen 7325 pounds and 7675 pounds a :P(73254X= 7675) 2 score for (x=7325) 2 = Roll 2-73257500 2 = -1-75 z score for (x=7675) 2 = d.cl 2= 7675 - 7500 100 normal distribution table, p (2<1.75) Hence P17325 5 345 7675) = P(-4.75<251.75)- - Plzs 1-75) —P(75-6-75). A Lonely Forom normal distribution at plz<-175) 70-7599 -6.0401-0.91981 Hence, probability of observing a weight between (5 57325 upounds and 2675 pounds = 0.9198 CamScanner

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