Answer:
The probability that the maximum safe-weight will be exceeded is 0.0455 or 4.55%.
Step-by-step explanation:
Given:
Maximum safe-weight of 37 cars = 82 tons
∴ Maximum safe-weight of 1 car (x) = 82 ÷ 37 = 2.22 tons (Unitary method)
Mean weight of 1 car (μ) = 2 tons
Standard deviation of 37 cars = 0.8 tons
So, standard deviation of 1 car is given as:
[tex]\sigma=\frac{0.8}{\sqrt{37}}=0.13[/tex]
Probability that maximum safe-weight is exceeded, P(x > 2.22) = ?
The sample is normally distributed (Assume)
Now, let us determine the z-score of the mean weight.
The z-score is given as:
[tex]z=\frac{x-\mu}{\sigma}\\\\z=\frac{2.22-2}{0.13}\\\\z=\frac{0.22}{0.13}=1.69[/tex]
Now, finding P(x > 2.22) is same as finding P(z > 1.69).
From the z-score table of normal distribution curve, the value of area under the curve for z < 1.69 is 0.9545.
But we need the area under the curve for z > 1.69.
So, we subtract from the total area. Total area is 1 or 100%.
So, [tex]P(z > 1.69) = 1 - P(z < 1.69)[/tex]
[tex]P(z>1.69)=1-0.9545=0.0455\ or\ 4.55\%[/tex]
Therefore, the probability that the maximum safe-weight will be exceeded is 0.0455 or 4.55%.
