Answer:
a) 0.0081975
b) 0.97259
Step-by-step explanation:
The line width for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer
We solve using z score formula
z = (x-μ)/σ, where
x is the raw score
μ is the population mean
σ is the population standard deviation.
a. What is the probability that a line width is greater than 0.62 micrometer?
z = 0.62 - 0.5/0.05
z = 2.4
Probability value from Z-Table:
P(x<0.62) = 0.9918
P(x>0.62) = 1 - P(x<0.62)
= 0.0081975
b. What is the probability that a line width is between 0.4 and 0.63 micrometer?
For 0.4
z = 0.4 - 0.5/0.05
= -2
Probability value from Z-Table:
P(x = 0.4) = 0.02275
For 0.63
z = 0.63 - 0.5/0.05
= 2.6
Probability value from Z-Table:
P(x = 0.63) = 0.99534
P(x = 0.63) - P(x = 0.4)
= 0.99534 - 0.02275
= 0.97259
c. The line width of 90% of samples is below what value?
Answer:
0.148 or 14.80%
Step-by-step explanation:
Mean overhead reach (μ) = 200 cm
Standard deviation (σ) = 8.9 cm
The z-score for any reach, X, is given by:
For a X = 209.30 cm:
A z-score of 1.045 corresponds to the 85.20th percentile.
Therefore, the probability that an individual distance is greater than 209.30 cm is:
The probability is 0.148 or 14.80%
Answer:
Step-by-step explanation:
u have a difference of squares
25x^2 - 16 = 0
(5x - 4)(5x + 4) = 0
5x - 4 = 0 5x + 4 = 0
5x = 4 5x = -4
x = 4/5 x = -4/5
so x = 4/5 or x = -4/5
-36 one of many possible answers.
