MATHS B – TERM 2 ASSIGNMENT
Simple Routine
QUESTION 1 | Wins | Tally | Frequency | Relative Frequency | i) | 4 | I | 1 | 0.00833 or 1120 | ii) | 3 | IIIII | 5 | 0.04166 or 5120 | iii) | 2 | IIIII IIIII IIIII IIIII IIIII I | 26 | 0.21666 or 26120 | iv) | 1 | IIIII IIIII IIIII IIIII IIIII IIIII IIIII IIIII IIIII III | 48 | 0.40000 or 48120 | v) | 0 | IIIII IIIII IIIII IIIII IIIII IIIII IIIII IIIII | 40 | 0.33333 or 40120 | a)

b) 450÷120=3.75 i) 1×3.75=3.75=3.75450=0.00833 ii) 5×3.75=18.75=18.75450=0.04166 iii) 26×3.75=97.5=97.5450=0.21666 iv) 48×3.75=180=180450=0.40000 v) 40×3.75=150=150450=0.33333
The above calculations show that the number of wins per person would increase by 3.75%, so all original calculations from part a) have been multiplied by this number. This does not change the relative frequency from part a), as shown in the calculations on the right, and also in the table on the right.
QUESTION 2 a) The Cadbury promotion shown in the assessment sheet is viewed as a binomial distribution. This is because it uses the Bernoulli trials on determining whether the promotion is a success or a failure, however, because there is a one-in-five chance of success, this contributes to allowing five Bernoulli trials to be conducted, with hopefully, 1 of the 5 trials succeeding.

b) There are several methods for finding whether or not the Cadbury promotion represents a binomial distribution. These are represented as Bernoulli trials:

i) The chance of getting heads in a coin toss (50/50). ii) The likelihood of rolling a 5 on a dice (1/6). iii) The likelihood of winning the top prize in the lottery, instead of any other category (1/9 roughly)
Simple Non-Routine
QUESTION 3 a) X | P (x) | Probability | = | 5 | 1 p5 q0 | 1×155×450 | 0.00032 | 4 | 5 p4 q1 | 5×154×451 | 0.0064 | 3 | 10 p3 q2 | 10×153×452 | 0.0512 | 2 | 10 p2 q3 | 10×152×452 | 0.2048 | 1 | 5 p1 q4 | 5×151×454 | 0.4096 | 0 | 1 p0 q5 | 1150×455 | 0.32768 |

Probability=Chance of an event = Number of successful outcomesNumber of outcomes possible

Pascal’s Triangle
Pascal’s Triangle

X | P (x) | Probability | = | 10 | 1 p10 q0 | 1×1510×450 | 0.0000001024 | 9 | 10 p9 q1 | 10×159×451 | 0.000004096 | 8 | 45 p8 q2 | 45×158×452 | 0.000073728 | 7 | 120 p7 q3 | 120×157×453 | 0.000786432 | 6 | 200 p6 q4 | 200×156×454 | 0.00524288 | 5 | 252 p5 q5 | 252×155×455 | 0.0264241152 | 4 | 200 p4 q6 | 200×154×456 | 0.08388608 | 3 | 120 p3 q7 | 120×153×457 | 0.201326592 | 2 | 45 p2 q8 | 45×152×458 | 0.301989888 | 1 | 10 p1 q9 | 10×151×459 | 0.268435456 | 0 | 1 p0 q10 | 1×150×4510 | 0.1073741824 | b)

QUESTION 4
The probability distribution graphs shown in the previous two questions take the shape of a bell curve. This shows that the shape of the curve remains the same, even if the number of trials is increased. However, by increasing the number of trials, the curve becomes steeper, but the tail of the curve straightens out and becomes longer. This has been proven after increasing the number of trials from 5 to 10, with evidence of this theory shown above.
Complex Routine
QUESTION 5 a) Number of Wins | Experimental Frequency | Theoretical Frequency | Percentage error | 5 | 0 | 0.00032 | -100.00000 | 4 | 0.008333333 | 0.0064 | 30.20833 | 3 | 0.041666667 | 0.0512 | -18.61979 | 2 | 0.216666667 | 0.2048 | 5.79427 | 1 | 0.4 | 0.4096 | -2.34375 | 0 | 0.333333333 | 0.32768 | 1.72526 |

b) The error percentage above shows that there is a difference between the theoretical and the experimental probability distributions. In this, it shows that the experimental frequency is -83% lower than that of the theoretical frequency. The negative figure shows that the experimental frequency is greater, so anything below this number will be shown as a negative. The reason why the experimental