Question

Many body measurements of people of the same sex and similar ages such as height and upper arm length follow a Normal distribution reasonably closely. Weights, on the other hand, are not Normally distributed. Suppose a survey includes the weights of a representative sample of 548 females in the United States aged 20-29. Suppose the mean of the weights was 161.55 pounds, and the standard deviation was 48.97 pounds. The histogram of the data includes a smooth curve representing an N(161.55, 48.97) distribution Number in sample 20 40 60 80 100 120 140 160 0 50 100 350 400 150 200 250 300 Weight in pounds To access the complete data set, click the link for your preferred software format Excel Minitab JMP SPSS TI R Mac-TXT PC-TXT CSV Crunchit! From the histogram, the Normal curve does not appear to follow the pattern in the histogram that closely. Because of this, the use of areas under the Normal curve may not provide a good approximation to weights in various intervals. You may find Table A useful.

Answer 1

Bolutions- + From a given information, À sex vey includes the colights of a. representative sample of 548 females in the wited states ages 80 - 29. That is n=548 + The histogram of the data includes a smooth Cure representing an NC 16155, 48.97) distribution. Given diagram; Number in esample O DO 40 60 80 100 120 140 180 50 100 150 200 250 300 350 400 Weight in pounds.

That is , n = 548 (a) Given that There were 13 females that weighted andet 100 pounds The percent of females aged 30 to 29 weighted under 100 pounds B; ino. of females , -548 Tot ad females 13 -0.0937 2.3712 *The percent of the NC161.55,48-97) distribution fis given below 100 is : P(x2100)=P(XA < 100-J47 =P(Z < 100–161-55) = P(2<. ) 48. 97 /

- Plz2-1-26) (from standa) d normal tabk, area to the left of -1:26) = 0.1038 [10-381 (b) Given that There were 33 females that weighted aver 250 pounds The percent of females aged po to 29 weighted over 250 pounds is , crnoiaf fernales) 33 sloval ferales) 548 = 0.0602 Total females 548 = 6:027 'the percent of the N(161-55,48-99) distribution is above 950 is

Plx> 350) = P1 X=# > 35034) = Plz > 350 401 = P(Z > 98.63.) = P(Z >181) = 1- P(ZL181) >88.45 48.97/ (from standard normal table, area to the left of 1981) el-0.9649 = 0.0351 - [3.518 (c) ->By gaserving the answers in parts (a) and (b), it is not a good idea to Summarize the distribution of weights by an NC 161,55,48-99) distribution

to No, because the predicted percentages are significantly different than the actual percentages. Ans-d