13.Data from the article \"The Osteological Paradox: Problems inferring Prehisto

Question

13.Data from the article "The Osteological Paradox: Problems inferring Prehistoric Health from Skeletal Samples" (Current Anthropology (1992):343-370) suggests that a reasonable model for the distribution of heights of 5-year old children (in centimeters) is N(100, 62) . Let the letter X represent the variable "height of 5-year old", and use this information to answer the following. Use 4 decimal places unless otherwise indicated.

(a) P(X > 89.2) =  

(b) P(X < 109.78) =  

(c) P(97 < X < 106) =  

(d) P(X < 85.6 or X > 111.4) =  

(e) The middle 80% of all heights of 5 year old children fall between  and  . (Use 2 decimal places.)

1.Many random number generators allow users to specify the range of the random numbers to be produced. Suppose that you specify that the random number Y can take any value between 0 and 2. Then the density curve of the outcomes has constant height between 0 and 2, and height 0 elsewhere.
(a) Is the random variable Y discrete or continuous? Why?
This is a continuous random variable because it has a finite sample space.This is a discrete random variable because it has a finite sample space.    This is a continuous random variable because the set of possible values is an interval.This is a discrete random variable because the set of possible values is an interval. [Incorrect: Your answer is incorrect.]
(b) What is the height of the density curve between 0 and 2? Draw a graph of the density curve.


(c) Use your graph from (b) and the fact that probability is area under the curve to find P(Y 1).
2.A density curve of a uniform distribution takes the constant value k over the interval from 0 to 1.5 and is zero outside that range of values. This means that data described by this distribution take values that are uniformly spread between 0 and 1.5.

What does the total area under this density curve have to be?


What is the value of k? (Round your answer to three decimal places.)


What percent of the observations lie above 1.0? (Round your answer to a whole number.)
%

What percent of the observations lie below 0.1? (Round your answer to a whole number.)
%

What percent of the observations lie between 0.1 and 1.0? (Round your answer to a whole number.)
%

Explanation / Answer

13) here std deviation=(62)1/2=7.874

also zscore=(X-mean)/std deviation

a) P(X>89.2)=1-P(X<89.2)=1-P(Z<(89.2-100)/7.874)=1-P(Z<-1.3716)=1-0.0851=0.9149

b)P(X<109.78)=P(Z<(109.78-100)/7.874)=P(Z<1.242)=0.8929

c)P(97<X<106)=P(-0.381<Z<0.762)=0.777-0.3516=0.4254

d)P(X<85.6 or X>111.4)=1-P(85.6<X<111.4)=1-P(-1.8288<Z<1.4478)=0.9262-0.0337=0.8924

e)value of z at 90 percentile z=1.6449

middle 80% value fall b/w =mean +/- z*std deviation=87.05 ; 112.95

2)a)This is a continuous random variable because the set of possible values is an interval

b) as area =1=sum of probability

height*base=1

therefore height =1/2=0.5

c) total area under this density curve have to be=1

k=1/1.5

P(X>1)=1-P(X<1)=1-1/1.5=0.3333 hence 33.33% values

P(X<1) =1/1.5=0.6667 hence 66.67%

P(0.1<X<1)=(1-0.1)/1.5=0.9/1.5=0.6 hence 60%

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