STAM4000-Quantitative Methods

STAM4000-Quantitative Methods, Assignment 1 - 2012 Trimester 2

Table of Contents

Question 13

Question 24

Q2, Part a4

Q2, Part b5

Q2, Part c6

Question 36

Q3, Part a6

Q3, Part b7

Question 48

Q4, Part a8

Q4, Part b9

Q4, Part c10

Q4, Part d11

Assignment 1 - 2012 Trimester 2

Question 1

Standard Deviation is a measure of central tendency which shows the variation of data from the mean. Standard deviation is also used to indicate the distance from the mean.

Our bell shaped score distribution has the following statistics:

Mean = 70

Standard Deviation= 10

A student Pat has a score which is 2 standard deviations above the mean.

Standard deviation= 10

2 Standard Deviation = 20

Pat's approximate Score = 70+20 = 90

It is an approximate as the scores vary and mean is an aggregate measure.

As it evident from the mean, Pat scored relatively higher from other students as his score was above two standard deviations. As the distribution is bell shaped, mean of 70 means that the about 50% of students had score above 70. It can be said that students who scored above Pat cannot be more than 5% of the population as in bell shaped distribution; as the area under the curve till positive 2nd standard deviation is usually around 95% of all the data. The population must be 5% of the students who scored the highest marks.

Question 2

N = 50

P = 0.04

Q2, Part a

The data represents the characteristics of Binomial Distribution due to the following reasons:

The number of mobile phones to be chosen “n” is fixed.

The question explicitly states that the selection of all phones is independent of the selection of other phones.

Each phone will either be a malfunctioning unit or not, which satisfies the tow outcomes arrival of samples selection.

As stated in the question the probability remains the same for all smart phones.

The probability mass function of binomial distribution is stated below:

F (k,n,p) = P (K= k) =

Here in the function:

k = is the number of malfunctioning units

N = 50

P = 0.04

When no phone will malfunction, k = 0.

The probability that no phone will malfunction will be given by

F (Function) (0:50,.04) = P (k=0) =

F (Function) (0:50,.04) = P (k=0) = (0.96)^50

F (Function) (0:50,.04) = P (k=0) = (1) (0.12988)

P (k=0) = 0.12988

So the probability of getting no malfunctioning phone is 0.1298 or approximately 13%.

Q2, Part b

Probability of having at most 2 malfunctioning phones is calculated by:

P (Probability) (k=2) = P (Probability) (k=0) + P (Probability) (k=1) + P (Probability) (k=2)

P (Probability) (k=0) = 0.12988

P (Probability) (k=1) =

P (Probability) (k=1) = 50 (0.04) (0.96)^49

P (Probability) (k=1) = 2 * (0.1352)

P (Probability) (k=1) = 0.27059

P (Probability) (k=2) =

P (Probability) (k=2) = (1225) (0.0016) (0.96)^48

P (Probability) (k=2) = 0.2762

Probability of having at most 2 malfunctioning phones is calculated by:

P (Probability) (k=2) = P (Probability) (k=0) + P (Probability) (k=1) + P (Probability) (k=2)

P (Probability) (k=2) = 0.12988+0.27059+0.2762

P (Probability) (k=2) = 0.67667

Probability of having at most 2 malfunctioning phones is P (Probability) (k=2) = 0.67667

Q2, Part c

The probability of no malfunction = 1-( P (k=0)) ...