Answer:
Step-by-step explanation:
Since the width of a tool used for semiconductor manufacturing is assumed to be normally distributed,
we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = width of selected tools.
µ = mean width
σ = standard deviation
From the information given,
µ = 0.5 micron
σ = 0.05 micron
We want to find the probability that a randomly selected tool will have a width between 0.47 and 0.63 microns. It is expressed as
P(0.47 ≤ x ≤ 0.63)
For x = 0.47,
z = (0.47 - 0.5)/0.05 = - 0.6
Looking at the normal distribution table, the probability corresponding to the z score is 0.27
For x = 0.63,
z = (0.63 - 0.5)/0.05 = 2.6
Looking at the normal distribution table, the probability corresponding to the z score is 0.9953
P(0.47 ≤ x ≤ 0.63) = 0.9953 - 0.27 = 0.7253
Second question
90% = 90/100 = 0.9
Looking at the normal distribution table, the z score corresponding to 90% is 1.285
Therefore
1.285 = (x - 0.5)/0.05
1.285 × 0.05 = x - 0.5
0.06425 = x - 0.5
x = 0.5 + 0.06425
x = 0.56425
The width value is approximately 0.56
