TASK A
1. Kristen has 150 red tiles, 250 blue tiles and 350 green tiles. She grabs a big handful of 50 tiles all at once and places them on a table where she counts to see how many of them are green.
a. Hypergeometric or Binomial: Hypergeometric
b. What is the probability that she pulled out exactly 21 green tiles? 0.092961
c. What is the probability that she pulled out at least 21 green tiles? 0.796748
2. Kawhi Leonard’s career field goal percentage is 0.475. In each round of the playoffs, he is expected to shoot around 160 shots. I am wondering how many of those shots will be successful.
a. Hypergeometric or Binomial: Binomial
b. What is the probability that he hits exactly 80 of those shots? 0.051551
c. What is the probability that he hits less than 70 of those shots? 0.151708
3. Caleb is conducting a survey in his old elementary school. The school has 650 students in total and he knows that exactly 14% of the students in the school are in grade 8. He decides to randomly survey 55 students in the school and is interested in knowing how many of them are in grade 8.
a. Hypergeometric or Binomial: Hypergeometric
b. What is the probability that exactly 9 of the students are in grade 8? 0.132527
c. What is the probability that more than 11 of the students are in grade 8? 0.066731
TASK B/C not defined?
NORMAL APPROXIMATION
Before applying this approximation, we need to ensure that our sample is less than one tenth of the population, otherwise the approximation will not be sufficiently accurate:
(1) 50/750=1/15 OK.
(2) The goal average 0.475 is taken over a whole career, consisting of many rounds, so we can assume that 1 round is a small percentage of that career.
(3) 55/650<1/10 OK.
Now need the mean and standard deviation:
(1) Mean=50×350/750=70/3 (23.33...), applying the same proportion to the sample (50 tiles) as is found in the population (350 green tiles divided by total tiles). For SD we need the failure fraction = (750-350)/750=1-350/750=8/15 (0.533...). We also need (N-n)/(N-1)=where N=750, n=50=100/107 (0.9345...)
SD=√((70/3)(8/15)(100/107))=3.41 approx.
(1b) To estimate the probability for a discrete value we need to apply a correction: 9 is equivalent to the range 20.5-21.5. We need Z scores for each (Z₁ and Z₂): Z₁=(20.5-(70/3))/3.41=-0.83 approx; Z₂=(21.5-(70/3))/3.41=-0.54 approx. Percentages are 0.203 (Z₁) and 0.295 (Z₂), the difference is 0.092, which compares favourably with the former figure of 0.093.
(1c) “At least 21” means the minimum is 20.5 after correction, so Z=-0.83 corresponding to a probability of 0.203 (for X≤21) therefore, for X≥21 we use 1-0.203=0.797, same as (1c) in Task A, after rounding.
(2) Mean=0.475×160=76; SD for Binomial is √(160(0.475)(1-0.475))=√((76)(0.525))=6.32 approx.
(2b) Range correction on 80 is 79.5-80.5. Z₁=(79.5-76)/6.32=0.554, Z₂=(80.5-76)/6.32=0.712, corresponding to 0.710 and 0.762. Therefore 0.762-0.710=0.052, same as (2b) in Task A, after rounding.
(2c) Correction for <70 is <69.5, Z=(69.5-76)/6.32=-1.029 corresponding to 0.152, same as (2c) in Task A, after rounding.
(3) Probability of Grade 8 students=0.14, corresponding to 0.14×650=91 students. Mean=0.14×55=7.7 for the sample of 55 students. SD=√((7.7)(0.86)(650-55)/649)=2.46. Range correction for 9 is 8.5-9.5.
Z₁=(8.5-7.7)/2.46=0.325, corresponding to 0.627; Z₂=(9.5-7.7)/2.46=0.731, corresponding to 0.767, so the approximation is 0.767-0.627=0.140 compared to 0.133 in Task A.
“More than 11” after correction is ≥11.5. Z=(11.5-7.7)/2.46=1.54, corresponding to 0.938. But this is ≤11.5, so for ≥11.5, we need 1-0.938=0.062, compared to 0.067 in Task A.