Question

Fifty randomly selected students were asked the number of movies they watched them previous week. The results are as follows:

# of Movies

# of Movies | Student Frequency

0. 10

1 . 18

2 . 12

3 . 8

4 . 2

a) Find the sample mean, median, and range of the sample. .

b) Find the standard deviation and the variance.

c) Find the first quartile. (1 mark)

d) Find the second quartile. To which value it corresponds? (1 mark)

e) Find the third quartile. (1 mark)

f) What percent of the students saw fewer than three movies.

j) Find the 40th percentile.

h) Find the 90th percentile.

Answer 1

Explanation :

Mean :

For a data set, the arithmetic mean, also called the expected value or average, is the central value of a discrete set of numbers: specifically, the sum of the values divided by the number of values.

Median :

In statistics and probability theory, a median is a value separating the higher half from the lower half of a data sample, a population or a probability distribution. For a data set, it may be thought of as "the middle" value.

Range :

The Range is the difference between the lowest and highest values.

Standard Deviation (SD ) :

The Standard Deviation is a measure of how spread out numbers are.

Its symbol is σ (the greek letter sigma)

The formula is easy: it is the square root of the Variance.

Variance :

The Variance is defined as the average of the squared differences from the Mean.

Steps to calculate the variance are as follows :

1. Work out the Mean (the simple average of the numbers)
2. Then for each number: subtract the Mean and square the result (the squared difference).
3. Then work out the average of those squared differences.

First Quartile :

The lower quartile, or first quartile, is denoted as Q1 and is the middle number that falls between the smallest value of the dataset and the median.

Second Quartile :

The second quartile, Q2, is also the median.

In the light of the above discussions, the solution to the first 4 subparts of the given sets of transactions are as follows:

The Solution :

Q. a.

The Mean = 1/50 [ 0*10 + 1*18 + 2*12 + 3*8 + 4*2 ] = 1/50 [ 0+18+24+24+8 ] = 1/50 [ 74 ] = 74/50 = 1.48

The Median = the middle value = 12

The Range = Highest value - Lowect value = 18 - 2 = 16

Q. b.

The SD^2 =  [ ∑f⋅M^2−n(μ)^2/(n−1) ] = 1/(n-1) [  ∑ f * ∑ M^2 - n(μ)^2 ]

= 1/(n-1) [ 50 * { 0^2 + 1^2 + 2^2 + 3^2 + 4^2 } - 50 ( 1.48 )^2 =  1/(n-1) [ 50* { 0 + 1 + 4 + 9 + 16 } - 50*2.1904

= 1/ ( 50 - 1 ) [ 50*30 - 109.52 } = 1/49 [ 1500 - 109.52 ] = 1/49 [ 1390.48 ] = 1390.48/49 = 28.3771428

Therefore, SD = root over ( 28.3771428 ) = 5.32702006 = 5.327 (app )

The Variance = SD^2 = 28.3771428

Q. c.

First Quartile =  middle number that falls between the smallest value of the dataset and the median = 18

Q. d.

Second Quartile = the middle value = 12 = the median

Therefore the Second Quartile is also the Median.

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