1) The outcomes for rolling two dice, the sample space, is as follows:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
There are 36 outcomes in the sample space.
2) The ways to roll an odd sum when rolling two dice are:
(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5). There are 18 outcomes in this event.
3) The probability of rolling an odd sum is 18/36 = 1/2 = 0.5
Answer:
Step-by-step explanation:
Since the number of pages that this new toner can print is normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = the number of pages.
µ = mean
σ = standard deviation
From the information given,
µ = 2300 pages
σ = 150 pages
1)
the probability that this toner can print more than 2100 pages is expressed as
P(x > 2100) = 1 - P(x ≤ 2100)
For x = 2100,
z = (2100 - 2300)/150 = - 1.33
Looking at the normal distribution table, the probability corresponding to the z score is 0.092
P(x > 2100) = 1 - 0.092 = 0.908
2) P(x < 2200)
z = (x - µ)/σ/√n
n = 10
z = (2200 - 2300)/150/√10
z = - 100/47.43 = - 2.12
Looking at the normal distribution table, the probability corresponding to the z score is 0.017
P(x < 2200) = 0.017
3) for underperforming toners, the z score corresponding to the probability value of 3%(0.03) is
- 1.88
Therefore,
- 1.88 = (x - 2300)/150
150 × - 1.88 = x - 2300
- 288 = x - 2300
x = - 288 + 2300
x = 2018
The threshold should be
x < 2018 pages
