# The following stem and leaf display shows the dailyy production of Blu-ray disc players.Stem and leaf display: Production Stem and leaf of text messages N=* Leaf Unit = 10   5        ...

The following stem and leaf display shows the dailyy production of Blu-ray disc players.

Stem and leaf display: Production

Stem and leaf of text messages N=*

Leaf Unit = 10

5         8        12334

15       8        556667899

30       9        11111233333344

(25)     9        5556667777778888888899999

25        10      00111223334444

11        10      5566678

4          11      2334

a) Determine the missing value (*). Explain or show how you arrived at your answer.

b) What is the smallest number of Blu-Ray disc players produced during a day?

c) During how many days was production less than 950 blu-ray disc players?

d) What proportion of the days was 1050 or more bLu-ray disc player produced?

e) Comment on the skewness of the distribution and explain what this indicates with regards to the production of blu-ray disc players.

f) Find the 70th percentile of teh number of blu-ray disc players produced and explain what this vlaue indicates.

g) Calculate the inter-quartile range for blu ray disc players produced. interpret your answer.

vixen999 | High School Teacher | (Level 2) Adjunct Educator

Posted on

This appears to be a stem and leaf plot produced by the Minitab software which is why it has an additional column at the beginning.  The stem is the second column and the leaves are the third column.

a)  Count the number of stems and there are 78, so N=78.

b)  The smallest number is the first stem and the first leaf.  It says that the leaves are each 10, so the smallest number if 8,1 so 810.

c)  The "9" stem has been split into two.  The second one starts at 950.  If you count the stems below that, you get 28 so 28 days had production below 950.

d)  Looking at the 2nd stem of "10," all of the entries in that row and the one after it are of values 1050 or greater.  There are 11 of these out of 78.  So, 11/78, or 14.1% of the days had production of 1050 or greater.

e)  The data is heavily skewed towards production within the range of 910 to 990.  This would indicate that perhaps a standard production goal is somewhere within that range or that the average demand for production is near there.  It appears that there are not many days that production is high -- over 1050.  So, either the equipment/labor cannot regularly produce this many Blu-Rays or the production schedule based on demand does not require this high level of production very often.  It could also be an example of days when extra "lines" are run or additional labor is used.

f)  In the first column, the number(25) indicates that the median (middle) value is in that row.  Averaging the 39th and 40th values gives us a median value of 985.  Normally, we would look at the upper and lower quartiles.  in this case, we want to look at the 70th percentile which is very close to the 75 percentile which represents the upper quartile.  70% of 78 would give us a value of 54.6, so we would look at the 54th and 55th nmber which are 1020 and 1030, so the average is 1025.  So, 70% of our production runs are of 1025 Blu-Ray discs or fewer.

g)  The upper quartile (75%) is between 1030 and 1040 which would give us an average of 1035.  Our lower quartile would be at the 18th and 19th values which are both 910.  So, our interquartile range is 1035 - 910 which gives us an interquartile range of 125.  This means that most of the production days are within the range of 910 to 1035 Blu-Ray players.  This seems to be a fairly large spread of 125/910 or 13% of the minimum value.

mellove | Student, Undergraduate | eNotes Newbie

Posted on

Please see the correction to the third data set.  there should have been three 4's to the end and not two

30       9        111112333333444