Question

Construct a frequency distribution and a relative frequency distribution for the light bulb data with a class width of 20, starting at 800. Copy and paste your distribution tables here. 2. Construct a histogram based on this frequency distribution table for the light bulb data. Copy and paste your histogram here. Describe the shape of the histogram. (Is it unimodal, bimodal, skewed, etc.?) skewed 3. Now, construct a frequency distribution table and a relative frequency distribution table for the light bulb data with a class width of 100, starting at 800. Copy and paste your distribution tables here. 5. Construct a histogram based on this frequency distribution table for the data. Copy and paste your histogram here. Describe the shape of the histogram. (Is it unimodal, bimodal, skewed, etc.?)
819

836

888

897

903

907

912

918

942

943

952

959

962

986

992

994

1004

1005

1007

1015

1016

1018

1020

1022

1034

1038

1072

1077

1077

1082

1096

1100

1113

1113

1116

1153

1154

1174

1188

1230

Answer 1

Answer:

Step-by-step explanation:

Hello!

You have the information about light bulbs (i believe is their lifespan in hours) And need to organize the information in a frequency table.

The first table will be with a class width of 20, starting with 800. This means that you have to organize all possible observations of X(lifespan of light bulbs) in a class interval with an amplitude of 20hs and then organize the information noting their absolute frequencies.

Example

1) [800;820) only one observation classifies for this interval x= 819, so f1: 1

2)[820; 840) only one observation classifies for this interval x= 836, so f2: 1

3)[840;860) no observations are included in this interval, so f3=0

etc... (see attachment)

[ means that the interval is closed and starts with that number

) means that the interval is open, the number is not included in it.

fi: absolute frequency

hi= fi/n: relative frequency

To graph the histogram you have to create the classmark for each interval:

x'= (Upper bond + Lower bond)/2

As you can see in the table, there are several intervals with no observed frequency, this distribution is not uniform least to say symmetric.

To check the symmetry of the distribution is it best to obtain the values of the mode, median and mean.

To see if this frequency distribution has one or more modes you have to identify the max absolute frequency and see how many intervals have it.

In this case, the maximal absolute frequency is fi=6 and only one interval has it [1000;1020)

$Mo= LB + Ai (\frac{D_1}{D_1+D_2} )$

LB= Lower bond of the modal interval

D₁= fmax - fi of the previous interval

D₂= fmax - fi of the following interval

Ai= amplitude of the modal interval

$Mo= 1000 + 20*(\frac{(6-3)}{(6-3)+(6-4)} )=1012$

This distribution is unimodal (Mo= 1012)

The Median for this frequency:

Position of the median= n/2 = 40/2= 20

The median is the 20th fi, using this information, the interval that contains the median is [1000;1020)

$Me= LB + Ai*[\frac{PosMe - F_{i-1}}{f_i} ]$

LB= Lower bond of the interval of the median

Ai= amplitude of the interval

F(i-1)= acumulated absolute frequency until the previous interval

fi= absolute frequency of the interval

$Me= 1000+ 20*[\frac{20-16}{6} ]= 1013.33$

Mean for a frequency distribution:

$X[bar]= \frac{sum x'*fi}{n}$

∑x'*fi= summatory of each class mark by the frequency of it's interval.

∑x'*fi= (810*1)+(230*1)+(870*0)+(890*2)+(910*4)+(930*0)+(950*4)+(970*1)+(990*3)+(1010*6)+(1030*4)+(1050*0)+(1070*3)+(1090*2)+(1110*4)+(1130*0)+(1150*2)+(1170*1)+(1190*1)+(1210*0)+(1230*1)= 40700

$X[bar]= \frac{40700}{40} = 1017.5$

Mo= 1012 < Me= 1013.33 < X[bar]= 1017.5

Looking only at the measurements of central tendency you could wrongly conclude that the distribution is symmetrical or slightly skewed to the right since the three values are included in the same interval but not the same number.

*-*-*

Now you have to do the same but changing the class with (interval amplitude) to 100, starting at 800

Example

1) [800;900) There are 4 observations that are included in this interval: 819, 836, 888, 897 , so f1=4

2)[900;1000) There are 12 observations that are included in this interval: 903, 907, 912, 918, 942, 943, 952, 959, 962, 986, 992, 994 , so f2= 12

etc...

As you can see this distribution is more uniform, increasing the amplitude of the intervals not only decreased the number of class intervals but now we observe that there are observed frequencies for all of them.

Mode:

The largest absolute frequency is f(3)=15, so the mode interval is [1000;1100)

Using the same formula as before:

$Mo= 1000 + 100*(\frac{(15-12)}{(15-12)+(15-8)} )=1030$

This distribution is unimodal.

Median:

Position of the median n/2= 40/2= 20

As before is the 20th observed frequency, this frequency is included in the interval [1000;1100)

$Me= 1000+ 100*[\frac{20-16}{15} ]= 1026.67$

Mean:

∑x'*fi= (850*4)+(950*12)+(1050*15)+(1150*8)+(1250*1)= 41000

$X[bar]= \frac{41000}{40} = 1025$

X[bar]= 1025 < Me= 1026.67 < Mo= 1030

The three values are included in the same interval, but seeing how the mean is less than the median and the mode, I would say this distribution is symmetrical or slightly skewed to the left.

I hope it helps!