Ex. 7 / A university investigated how much space on its computers’ hard drives is used for data storage. The results are shown below. It is given that 40 hard drives use less than 20GB for data storage.

a. Find the total number of hard drives represented.

If 40 hard drives use less than 20GB, it means that in the interval of 0-20 there are 40 drives (2 cube=40 drives => 1 cube=20 drives)

There are 24 cubes in the whole histogram, which means that there are in total 24*20=480 hard drives.

b. Calculate an estimate of the number of hard drives that use less than 50GB.

An estimate of the number of the number of hard drives that use less than 50GB equals 130 (interval of 0-20 + interval of 20-50 | 40+60+60/2=40+60+30=130)

c. Estimate the value of k, if 25% of the hard drives use kGB or more.

The number of drives in the interval of 120-200 is 80, and it’s the 1/6 of the total number, which is 480. We add 40 (80/2) to get 25% of 480, therefore 80+40=120 hard drives (which is the 1/4 or 25% of 480).

Ex. 9 / The thicknesses, k mm, of some steel sheets are represented in the histogram. It is given that k < 0.4 for 180 sheets.

a. Find the ratio between the frequencies of the three classes. Give your answer in simplified form.

The ratio between the frequencies of the 3 classes:

1st class: (0,4-0,1)*4=0,3*4=1,2
2nd class: (0,8-0,4)*2=0,4*2=0,8
3rd class: (1,1-0,8)*1=0,3

The ratio is 12:8:3

b. Find the value of n, given that frequency density measures sheets per n mm.

1,2/n=180 => n=1/150

c. Calculate an estimate of the number of sheets for which:

i. k < 0.5;

The number of sheets in interval of 0,2-0,4=180
180+(120/4)=180+30=210

We divide by 4 because 0,4-0,5 interval is 1/4 of whole 0,4-0,8 interval (25% of it).

ii. 0.75 ≤ k < 0.94

(0,05*2*150)+(0,4*1*150)=14+21=36

i. = 210; ii. = 36

d. The sheets are classified as thin, medium or thick in the ratio 1:3:1. Estimate the thickness of a medium sheet, giving your answer in the form a ≤ k < b. How accurate are your values for a and b?

The 1st class = 0,1 ≤ k < 0,4 (considered as a); the 2nd class = 0,4 ≤ k < 0,8 (considered as b).

(0,4-a)*4*150=180-69 (sheets)=> a=0,215
(b-0,4)*2*150=208-(180-69)=> b=0,720

The thickness of medium sheet is 0,215 ≤ k < 0,720 mm