A ferry will safely accommodate 82 tons of passenger cars. Assume that the meanweight of a passenger car is 2 tons with standard deviation 0.8 tons. If a random sampleof 37 cars are loaded onto the ferry, what is the probability that the maximum safeweight will be exceeded?

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:

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:

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,

Therefore, the probability that the maximum safe-weight will be exceeded is 0.0455 or 4.55%.

i think it is 487.5. i may be wrong tho. hope this helps!