FINAL Problem 3 Your body mass index (BMI) is your weight in kilograms divided b

Question

FINAL Problem 3 Your body mass index (BMI) is your weight in kilograms divided by the square of your height in meters. According to Vital and Health Statistics, No. 361, 2005 of National Center for Health Statistics (http://www.cdc.gov/nchs/data/ad/ad361.pdf, page 20), the BMIs of American women (ages 20 to 29 in 1999-2002) are approximately normally distributed with a mean of 26.8 kg/m2 and a standard deviation of 7.4 kg/m2.

Q1. (5 points) If people with BMIs over 25kg/m2 are over weighted or obese, what is the probability that a randomly selected American woman of ages 20 to 29 is over weighted or obese? Ans. P[X>25]=P[(X-26.8)/7.4>(25-26.8)/7.4] = P[Z>-0.24] = 0.5948

Q2. (5 points) If people with BMIs between 18.5 and 24.9kg/m2 inclusively are normally weighted, what is the probability that a randomly selected American woman of ages 20 to 29 is normally weighted? Ans. P[18.5<=X<=24.9]=P[-1.12<=Z<=-0.26]= 0.3974-0.1314=0.266

Q3. (5 points) If the probability that a randomly selected American woman of ages 20 to 29 is under weighted is 0.1314 and the probability that a randomly selected American 2 woman of ages 20 to 29 is obese is 0.3336, what are the maximum BMI for under- weighted American women of ages 20 to 29 and the minimum BMI for obese American women of ages 20 to 29, respectively? Ans. 26.8-1.12*7.4 = 18.512; 26.8+0.43*7.4 = 29.982

Q4. (10 points) Suppose that BMIs of American women of ages 30 to 39 also follow a normal distribution approximately but with different mean and standard deviation. It has been found that 5% American women of ages 30 to 39 have BMIs less than or equal to 19.0 kg/m2 and 10% have BMIs greater than or equal to 38.0 kg/m2. What are the mean and standard deviation of the BMI distribution for American women of ages 30 to 39? Ans.19.0- =-1.645,38.0- =1.28, = (19.0*1.28+1.645*38.0)/2.925 = 29.68547, = (38.0-(19.0*1.28+1.645*38.0)/2.925)/1.28 = 6.495726 or 19.0- =-1.64,38.0- =1.28, = (19.0*1.28+1.64*38.0)/2.92 = 29.67123 = (38.0-(19.0*1.28+1.64*38.0)/2.92)/1.28 = 6.506849 or 19.0- =-1.65,38.0- =1.28, = (19.0*1.28+1.65*38.0)/2.93 = 29.69966 = (38.0-(19.0*1.28+1.65*38.0)/2.93)/1.28 = 6.484642

Please show how to get each of these answers. Thank you so much!

Explanation / Answer

1 ans) given mean =26.8

standard deviation = 7.4

x^=25

we know that standard z formulea

z= (x^- mean) / standard deviation = (25-26.8)/7.4 =-0.24

now from standard z table look into -0.24 value we have corresponding value is

the probability that a randomly selected American woman of ages 20 to 29 is over weighted or obese

P[Z>-0.24] = =1- 0.4052= 0.5948

2)

  given mean =26.8

standard deviation = 7.4

x1^=18.5

x2^=24.9

we know that standard z formulea

z1= (x^- mean) / standard deviation =(18.5-26.8) /7.4 = -1.12

now from standard z table look into -1.12 value we have corresponding value is 0.1314

z2= =(x^- mean) / standard deviation =(24.9-26.8) /7.4 = -0.26

now from standard z table look into -0.26 value we have corresponding value is 0.3974

P[-1.12<Z<-0.26]= 0.3974-0.1314=0.266

3)ans maximum and minimum

are given by

X^+- Z* SD

26.8-1.12*7.4 = 18.512

26.8+0.43*7.4 = 29.982

4) 19.0- =-1.645,

38.0- =1.28,

= (19.0*1.28+1.645*38.0)/2.925 = 29.68547,

= (38.0-(19.0*1.28+1.645*38.0)/2.925)/1.28 = 6.495726

or 19.0- =-1.64,

38.0- =1.28,

= (19.0*1.28+1.64*38.0)/2.92 = 29.67123

= (38.0-(19.0*1.28+1.64*38.0)/2.92)/1.28 = 6.506849

or 19.0- =-1.65,38.0- =1.28,

= (19.0*1.28+1.65*38.0)/2.93 = 29.69966

= (38.0-(19.0*1.28+1.65*38.0)/2.93)/1.28 = 6.484642

.

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