Answer:
The probability that defect length is at most 20 mm (or less than 20 mm) is 0.0505
Step-by-step explanation:
Mean defect length = u = 32
Standard deviation = [tex]\sigma[/tex] = 7.3
The distribution is normal and we have the value of population standard deviation so we will use the concept of z-score and probability from z-table to find the said probability.
We have to find the probability that the defect length is at most 20 mm. At most 20 mm means equal to or lesser than 20 mm. Since the normal distribution is a continuous distribution, the probability of at most is approximately equal to probability of lesser than.
If X represents the distribution of defect lengths, then we can write:
P(X ≤ 20) ≅ P(X < 20)
We can find the probability of defect length being lesser than 20 mm by converting it to z-score and find the corresponding probability from the z-table.
The formula to calculate the z-score is:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Substituting the values, we get:
[tex]z=\frac{20-32}{7.3}=-1.64[/tex]
Therefore, P(X < 20) is equivalent to P(z < -1.64)
From z-table we can find the probability of z being lesser than - 1.64, which comes out to be:
P(z < -1.64) = 0.0505
Therefore, we can conclude that:
The probability that defect length is at most 20 mm (or less than 20 mm) is 0.0505
