PART 1 The weights of all one hundred (100) 9th graders at a school are measured

Question

PART 1

The weights of all one hundred (100) 9th graders at a school are measured, and it is found that the mean of all the measurements is 100 lbs., with a standard deviation of 15 lbs. Explain how you would use this information to determine the percentage of students who weighed between 85 lbs and 115 lbs. Make sure your explanation includes the use of z-scores.

PART 2

20 points Using the example in part 1 above, answer the following questions in the same document you used to answer the questions from part 1. Make sure you show all the work required by the questions. • What is the percentage of students who weighed between 85 lbs and 115 lbs? (Show calculations) • What is the z-score of a student who weighed 105 lbs? (Show calculations) • If a student weighed only 60 lbs, how many students out of the one hundred 9th graders would weigh more? (Hint: calculate the percentage 1st)

PART 3

20 points The following year, the 9th grade students are weighed again, and this time it is found that the mean of all weights is 108 lbs, and the standard deviation is 17 lbs. Answer the following questions, using the same document you used in parts 1 and 2 to show how you arrived at your answers. Compare two students, one from the first class who weighed 98 lbs, and one from the second class who weighed 100 lbs. • Which of the two students was heavier relative to their class? (Hint:compare z-scores) • What percentage of students in the first class were heavier than the 98 lb. student? • What percentage of students in the second class were heavier than the 100 lb. student?

I got part one, I need help with part 2 and 3

x1 = 85; x2 = 115; = 100; = 15
z = ( x1 - ) / = -1
z = ( x2 - ) / = 1
P (85 < x < 115) = P (-1 < z < 1) = 0.6827

Explanation / Answer

2)percentage of students who weighed between 85 lbs and 115 lbs=P(85<X<115)=P((85-100)/15<Z<(115-100)/15)

=P(-1<Z<1) =84.13%-15.87% =68.27%

z-score of a student who weighed 105 lbs =(105-100)/15 =1

for P(X>60)=P(Z>(60-100)/15) =P(Z>-2.6667)=99.62%

hence students out of the one hundred 9th graders would weigh more =100*99.62% =99.62

3)

for first class ; for 98lbs z score =(98-100)/15 =-0.1333

for second class; for 100 lbs z score =(100-108)/17 =-0.4706

as z score for 98lbs is higher threfore it was heavier relative to other class

percentage of students in the first class were heavier than the 98 lb. student=P(X>98)=P(Z>-0.1333)=44.70%

percentage of students in the second class were heavier than the 100 lb. student=P(X>100)=P(Z>-0.4706)=31.90%

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