Question

PRE-LAB EXERCISES average size. If d is the diameter in millimeters, some pebbles will be bigger and some smaller than the desired average do. Suppose you have a rock crusher. The pebbles it makes have different sizes, but you can adjust the 1. (a) In the histogram below, about how many pebbles are shown to have diameter of greater 150 100 (b) In this distribution (same histogram,) approximately how many pebbles are represented 50 altogether, including all sizes? 20 40 60 80 Pebble Diameter d in Millimeters

Need help with 1: b, c, d, e and f. And 2. Thanks!

Answer 1

1.

(b)

Number of pebbles represented by the given histogram (approximately) are: 10+20+40+90+110+140+145+140+110+100+50+35+15+10 =1,015​​​​​​ pebbles.

(c)

Given: X =65 mm.

Frequency table (including the relative frequency and cumulative relative frequency) of the above histogram is:

$\Sigma f$ =1015; Relative frequency =f/$\Sigma f$; Cumulative relative frequency = Total relative frequency upto that class ($\leq$ upper limit of the respective class).

From the 'Cumulative relative frequency' column of the above table (shown in Yellow shading):

P(X$\leq$65 mm) =0.793103448 (here P is the Probability).

Now, P(X>65 mm) =1 - P(X$\leq$65 mm) = 1 - 0.793103448 =0.206896552

Hence, the number of pebbles out of 7x10⁷ pebbles that would have diameter of grater than 65 mm =

7x10⁷​​x0.206896552 =14482759 (rounded off).

(d)

Mean of given set of numbers =(1+2+3+4+5+15)/6 =30/6 =5

The mean value of 5 is not in the middle because middle values are 3 and 4, that is, median =(3+4)/2 =3.5

(e)

​​​​Sample mean of the given data, $\bar{X} =$$\Sigma X/n =90.047/12$ =7.504 mm

Sample Std.deviation, s =$\sqrt{\Sigma( X-\bar{X})^2/(n-1)}=$2.841 mm

(Here, we use '(n-1)' instead of 'n' because it removes any bias).

(f)

On an average, the size of a pebble is 7.504 mm.

On an average, the deviation of a pebble from the average value of 7.504 mm is 2.841 mm. (that is, the average deviation of a pebble from its mean size is 2.841 mm and it may be less than or greater than the mean).