Answer:
P(X>25)=0.107488
P(20<X<30)= 0.440427
0r 0.53490
Explanation :
Given from the question that the percentage of defective chips produced in the manufacturing of semiconductor is 2%
We are to Assume that the chips are independent and that a lot contains 1000 chips.
Then the following probabilities are to be estimated
a)More than 25 chips are defective
b)Between 20 and 30 chips are defective.
The percentage of defective chips produced in the manufacturing of semiconductor is 2%, therefore, our p=0.02.
The total number of chips, hence our n=1000.
Let X denote the number of defective chips in the manufacturing of semiconductor. So we will be able to calculate our mean
So the mean of X,
np= (10000.02). = 20
The variance of X can also be calculated as
np(1-p)= (10000.020.98). = 19.6
nq= n(1-p)=1000(1-0.2)=980 >5
Since np>5 and n(1-p)>5, then the requirements are satisfied
a) the z- score which is the value decreased by the mean value np and divided by the standard deviation now = np(1-p)
Z=
can be determined here
Z= x-np/√np(1-p)
z= [25.5-1000(0.02)]/√[1000(0.02)(1-0.02)]
z=1.24
if we check the normal distribution table using P(X>25)= P(X>25.5) = P(z>1.24) then we have the value of 0.107488
b) z= 20.5-1000(0.2)/√1000(0.02)(1-0.02)
z= 0.11
z= x-np/√np-(1-np)
z= 29.5-1000(0.2)/√1000(0.02)(1-0.02) = 2.15
CHECK THE ATTACHMENT TO COMPLETE THE SOLUTION
Probabilities are used to determine the chances of events
The given parameters are:
Start by calculating the mean and the standard deviation
[tex]\bar x = np[/tex]
[tex]\bar x = 2\% \times 1000[/tex]
[tex]\bar x = 20[/tex] --- the mean
[tex]\sigma = \sqrt{\bar x \times (1 -p)}[/tex]
[tex]\sigma = \sqrt{20 \times (1 -2\%)}[/tex]
[tex]\sigma = 4.43[/tex] ---the standard deviation
(a) The probability that more than 25 chips are defective
Using continuity correction, the probability is represented as:
[tex]P(x > 25) = P(x > 25.5)[/tex]
Start by calculating the z-score using:
[tex]z = \frac{x - \bar x}{\sigma}[/tex]
So, we have:
[tex]z = \frac{25.5 - 20}{4.43}[/tex]
[tex]z = \frac{5.5}{4.43}[/tex]
[tex]z = 1.241[/tex]
The probability is then represented as:
[tex]P(x > 25) = P(z > 1.241)[/tex]
Using z probability calculator, we have:
[tex]P(x > 25) = 0.1073[/tex]
Hence, the probability that more than 25 chips are defective is 0.1073
(b) The probability that between 20 and 30 chips are defective
Using continuity correction, the probability is represented as:
[tex]P(20 < x < 30) = P(20.5 < x < 29.5)[/tex]
Start by calculating the z-scores for both x values:
[tex]z = \frac{x - \bar x}{\sigma}[/tex]
So, we have:
[tex]z = \frac{20.5 - 20}{4.43}[/tex]
[tex]z = \frac{0.5}{4.43}[/tex]
[tex]z = 0.113[/tex]
[tex]z = \frac{29.5 - 20}{4.43}[/tex]
[tex]z = \frac{9.5}{4.43}[/tex]
[tex]z = 2.144[/tex]
The probability is then represented as:
[tex]P(20 < x < 30) = P(0.113 < z < 2.144)[/tex]
Rewrite as:
[tex]P(20 < x < 30) =P( z < 2.144) - P(z < 0.113 )[/tex]
Using z probability calculator, we have:
[tex]P(20 < x < 30) =0.98398 - 0.54498[/tex]
[tex]P(20 < x < 30) =0.4390[/tex]
Hence, the probability that between 20 and 30 chips are defective is 0.4390
Read more about probabilities at:
https://brainly.com/question/15246027
