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  • 1 Post By billymac00
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February 6th, 2019, 11:34 AM   #1
Joined: Nov 2012
From: The Netherlands

Posts: 40
Thanks: 2

2 Issues on Stats

Issue 1:
The number of Late-comers(at school) is kept up to date at the "Ramson" school community. On a certain day, 38 class-8 students were too late. On the same day, 10 pupils from class-12 were also too late. All students together make up 26% of all pupils of the " Ramson " school. And This is 7% for class-12 students.
In which of these two groups of pupils (class-8 <-> class-12) are there relatively the most latecomers?

what I have tried so far is :

26%/100% X 38%= 9,9 % class-8 were too late
7%/100% X 10= 0,7% class-12 were too late

So relatively speaking, most latecomers are in class-8 , because the percentage is the highest.


This was marked wrong by my teachers . so I would like to know the why , and the right stats approach here please.

Issue 2 :
If a new-build home is delivered, inspection(of home) usually takes place. In addition, defects often come to light. In the year 2000 the Home society took a sample of 325 new homes. The results are shown in the following table 2:

*Please see attached picture file 1

At these 325 houses, the average number of defects per home turned out to be 28.6.
Instead of the average, the central tendency median is taken into account . Assume that the number of dwellings per class is evenly distributed (except of course in the last class).

a) Investigate whether the median is larger or smaller than the average.
If a new house has one or more defects, the builder gets two weeks to repair it. This often proves impossible. In research, only 94 of the 325 homes were in order after two weeks. The other 231 homes still showed a number of shortcomings. Of the 231 dwellings that still showed defects after two weeks, the data on the number of defects per dwelling are included in the cumulative frequency polygon of Chart 2. (The same classification was used as in Table 2.)

*Please see attached picture file 2

b) Which of these box plots fits best with the data of graph 2 ? explain your answer.

Because after two weeks only a small part of the 325 homes is in order, it seems like the builders are performing badly. But the 231 homes that were not yet in order now had about 8.9 defects on average. It follows that the builders repaired more than three-quarters of all defects in the two-week recovery period.

c) Show this with help of a calculation.


My solution so far:

a) Mediaan 325+1:2= 163 rd number
rest I do not have a clue

b) Answer B, because then visually to see that both Q1 and Q3 are equally far from the median.

c) No idea

Once again I would like to know the why , and the right approach here please.

Thanks for any help or pointers for the above Issues .
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hello_math is offline  
February 6th, 2019, 03:34 PM   #2
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Joined: Aug 2008
From: Blacksburg VA USA

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Math Focus: primes of course
I'll only address Issue 1.

ok, class-8 latecomers were 8 of 38. How many class-12 students at most would make class-12 have the higher proportion? If you set up a simple ratio you can see the answer is 47.
Now you just need to compute the actual class-12 students. Call it x.
We are given (38+x) =0.26*total and x=0.07*total.
Rearrange as total= (38+x)/0.26, total=x/0.07
So (38+x)/0.26 = x/0.07
This is one equation with one unknown. Solve for x. If x>47, class-8 has proportionally more latecomers, otherwise class-12 does...
Thanks from hello_math
billymac00 is offline  
February 8th, 2019, 02:38 AM   #3
Joined: Nov 2012
From: The Netherlands

Posts: 40
Thanks: 2

Hi All

This was our (my son''s) solution to Issue 2 (with the broken homes )

++++++++++++++++++++++++++++++++++++++++++++++++++ +++++++++++++++
a) There are total of 325 houses. So median number of defects would be the number of defects on the (326/2)th house, i.e. the 163rd house after arranging in increasing order of defects. This house will be in the group (21 to 30 defects). That group has houses from rank 113 to 200. Assuming even distribution in this group,
the median = (163 – 113)/(200-113) * 10 + 21 = 26.747
Since the median is a whole number in this case, we can approximately say median = 27 defects

b) C is the best box plot. In a box plot, the box is determined by 25th and 75th percentiles, with the line being the median. The lines extending outward indicate the range of the data set.
In the given scenario, median is less than 5, 25th percentile is less than 5 , and 75th percentile is around 10. The range of the data is between 0 and more than 80. So the max range is much more than the 75th percentile and the min range, 25th , median and 75th percentile are all close together. So C is the best option.

c) Assuming distribution in each group is uniform,
The total number of defects = summing over classes (a,b) : (a+b)/2 * number of houses in (a,b)
= 3*148 + 8*(182-14 + 15.5*(208-182) + 25.5*(217-20 +35.5*(223-217) +45.5*(227-223) + 55.5*(228-227) + 73.5*(231-22
= 2019

Average number of defects = Total / Number of houses = 8.77 approx

Number of defects before repair approximately = average of class * number of houses in that class summed over all classes.
=3*5 + 8*21 + 15.5*85 + 25.5*88 + 35.5*59 + 45.5*47 + 55.5*7 + 61*13
= 9159 defects

Number of defects after 2 weeks = 2019
Number of defects repaired = 9159 – 2019 = 7140
Percentage of defects repaired = 7140 / 9159 *100 = 77.9 %

So more than 75% of defects were repaired in 2 weeks.

++++++++++++++++++++++++++++++++++++++++++++++++++ ++++++++++

could anyone maybe test this for us (him ) ? so that he knows what he has done here is indeed making sense before he submits the assignment. This is not at all urgent and it is just a check-check double-check issue for us.
Thanks all !

*PS: the closed brackets are coming up as emojis....just fyi
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