Respuesta :
Answer:
0.49926
Step-by-step explanation:
We solve for this using z score formula
z-score is
z = (x-μ)/σ,
where x is the raw score
μ is the population mean
σ is the population standard deviation
The probability that defect length is at most 20 mm is calculated as:
x = 20mm, μ = 33 mm, σ = 7.9mm
z = 20 - 33/7.9
= -13/7.9
= -1.64557
Obtaining the Probability value from Z-Table:
Probability (At most 20mm) = P(x ≤ 20mm) P(z = -1.64557)
P(x ≤ 20) = 0.049926
Therefore, the probability that defect length is at most 20 mm is 0.049926
Answer:
The probability that defect length is at most 20 mm is 0.0495.
Step-by-step explanation:
We are given that the defect length of a corrosion defect in a pressurized steel pipe is normally distributed with a mean value of 33 mm and a standard deviation of 7.9 mm.
Let X = the defect length of a corrosion defect in a pressurized steel pipe
The z-score probability distribution for the normal distribution is given by;
               Z  =  [tex]\frac{X-\mu}{\sigma}[/tex]  ~ N(0,1)
where, [tex]\mu[/tex] = mean defect length = 33 mm
      [tex]\sigma[/tex] = standard deviation = 7.9 mm
So, X ~ Normal([tex]\mu=33 \text{ mm}, \sigma^{2} = 7.9^{2} \text{ mm}[/tex])
Now, the probability that defect length is at most 20 mm is given by = P(X [tex]\leq[/tex] 20 mm)
    P(X [tex]\leq[/tex] 20 mm) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{20-33}{7.9}[/tex] ) = P(Z [tex]\leq[/tex] -1.65) = 1 - P(Z < 1.65)
                               = 1 - 0.9505 = 0.0495       Â
The above probability is calculated by looking at the value of x = 1.65 in the z table which has an area of 0.9505.
