a)
Stem - and – leaf diagram for battery life, in hour
Stems leaves 0 4 2 4 5 6 8 3 0 2 8 8 4 0 5 6
The steam- and – leaf indicates the battery life, in hours, which was taken from 12 models of mobile phones.
As can be seen from diagram, the most significant was from 24 hours to 38 hours, which mass up eight models of mobile phone. “4 hours” was considered as the lowest battery life for 12 models.
b) The amount (in mg) of a certain mineral found in a sample of 13 types of vegetable:
(i) 264 266 298 317 342 426 451 492 512 545 562 631 1,049
The mean is: (264 + 266 + 298 +…+ 631 + 1,049) ÷ 13 = 473.46
Median position: (13+1) ÷ 2 = 7
Therefore, the median is 451.
(ii) The shape of the distribution based on the sample data is right-skewed. This is because the median (451) is smaller than the mean (473.46).
Question 2:
a)
i. Probability of no camera is defective:
P(0) = nCX πx(1-π)n-X = 20C0×0.050×(1-0.05)20-0= 1×1×0.358= 0.358 ii. Probability that at most 2 cameras are defective:
P(1) = nCX πx(1-π)n-X = 20C1×0.051×(1-0.05)20-1= 20×0.05×0.377=0.377
P(2) = nCX πx(1-π)n-X = 20C2×0.052×(1-0.05)20-2= 190×0.0025×0.397=0.189
P (at most 2) = P (0) +P (1) +P (2) = 0.358 + 0.377 + 0.189 = 0.924 iii. Mean: µ = n × π = 20 × 0.05 = 1 Standard deviation: n × π(1- π) = 0.95 = 0.975 b) i. Probability that exactly 12 cups of coffee were sold in 2 hours: P (12) = e-λ λx ÷ X! = e-202012 ÷ 12! = 0.0176 ii. Probability that in an hour more than 3 cups of coffee are sold: P (0) = e-10100 ÷ 0! = 0.0000454 P (1) = e-10 101 ÷1! = 0.000454 P (2) = e-10102 ÷2! = 0.00227 P (3) = e-10103