Distribution

Question

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

Answer 1

Part 1:

Yes, you change the statistic in question into
the z-score using the mean and standard deviation.

What happens next depends on the problem. Typically,
you have to manipulate the probability equation so
that you can use the Normal distribution table.
The normal distribution table does in fact list
probabilities in terms of the z-score.

From there you can find the probability in question.


Part 2:

Prob ( 85 <= X <= 115) =

Prob ( (85-100)/15 <= z < = (115-100)/15 )

Prob( -15/15 <= z <= 15/15) =

Prob ( -1 <= Z <= 1) =

Prob ( z <= 1) - Prob ( z<= -1) =

.8413 - 0.1587 <--- from the normal table

= .6286 = 62.86%



For x = 105, the z-score is (105-100)/15 = 5/15 = 1/3 = 0.3333....

Prob( X > 60) = 1 - Prob ( X<=60)

= 1 - Prob ( z <= (60-100)/15)

= 1 - Prob( z <= -40/15)

= 1 - Prob ( z <= -8/3)

= 1 - Prob ( z <= -2.666666...)

= 1 - 0.00387

= 0.99613 = 99.613%


Part 3:

The z-score for the 98 lb student from the first class
is Z_old = (98-115)/15 = -17/15 = -1.133333333


The z-score for the 100 lb student from the new class
is Z_new = (100-108)/17 - -8/17 = -0.4705882353...


So the second student is heavier relative to
the class average.


Prob ( X_old > 98) = 1 - Prob ( X_old <= 98)
= 1 - Prob ( Z_old < -1.133333....)
= 1 - 1292
= 0.8708 = 87.08%

Prob ( X_new > 100) = 1 - Prob ( X_new <= 100)
= 1 - Propb ( Z_new < -0.4705882353..)
= 1 - .3192
= .6808 = 68.08%