Question

17 The histogram shows information about the times taken by some students to finish a puzzle. 0.8 0.6 Frequency density 0.4 0.2 10 20 30 50 Time taken (minutes) (a) Complete the frequency table for this information. Time taken ( minutes) Frequency 4 25 < 30 5 30< <50 .) Find ona est extimate ior the lowo finsh the puzzeCouud you answe quatne of the tines taken to flnish puzzle Toku freguency=23 why do ue divi de 23 by 4?

Answer 1

b)

The lower quartile is the number that accumulates 1/4 (or 25%) of the total area under the density curve.

The histogram is not a density curve, but it's the closest to a density curve  for grouped data.

Since we have grouped data we can't calculate the exact value of the quartile

but we can find an estimation.

We want to find a value that accumulates 25% of the frequency.

In this case the "area" of your frequency density, which is the sum of the frequencies

is equal to 4+4+6+5+4= 23.

So you have to divide 23 by 4 to know how much is 25% of the total "area".

We have that 23/4= 5.75

So the first quartile Q1 must be in the second class because

the frequency for the first class is 4 and for the second class is also 4

and we have 4<5.75 <8.

Thus 5<Q1<15

Once we have found the class the quartile must be, we find the estimate by interpolating using

the following formula.

- CJi-1 Q1 = lower limitit ( ) (upper limit,-lower limit.) fi

where i is the class the quartile must be.

f_i = the frequncy of the class i

cf_i-1 is the cummulative frequency until(including) class i-1

Lower limit i =   lower limit of the class i

Upper limit i = upperlimit of the class i

Therefore

5.75 - 4 Q1-5+((15-5) 9.375

Answer: 9.375