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
To find the specification limit such that only 0.5% of the bulbs will not exceed this limit we proceed as follows;
From the z-table, a z-score of -2.57 cuts off 0.005 in the left tail; given the formula for z-score
(x-μ)/σ
we shall have:
(x-5000)/50=-2.57
solving for x we get:
x-5000=-128.5
x=-128.5+5000
x=4871.50
16
x * 3 = 48
so divide 48 by 3
which is 16 so x = 16
