Question

A ferry will safely accommodate 68 tons of passenger cars. Assume that the mean weight of a passenger car is 1.8 tons with standard deviation 0.5 tons. If a random sample of 35 cars are loaded onto the ferry, what is the probability that the maximum safe weight will be exceeded?

Answer 1

Answer:

$P (Z<66.4)=1$

Step-by-step explanation:

We want the probability that the 35 cars are loaded onto the ferry. Therefore:

Since  

μ=1.8  and  σ=0.5  we have:

P(X<35 )=P ( X−μ<35−1.8 )=P((X−μ)/σ<(35−1.8)/0.5)

Since  

(x−μ)/σ=Z and  

$(35-1.8)/0.5=66.4$

we have:

$P (X<35)=P (Z<66.4)$

Use the standard normal table to conclude that:

$P (Z<66.4)=1$

Answer 2

Answer:

P(Z > 1.692) = 0.0453

Therefore the probability is 0.045 or 4.5%

Step-by-step explanation:

Given;

Mean u = 1.8 tons

Standard deviation r = 0.5

Number of data n = 35

For the total weight of 35 cars to exceed 68 tons. The average weight of each car should exceed

X = 68/35 = 1.943

We can then use that to estimate the z value.

Z = (X - u)/(r/√n)

Z = (1.943 - 1.8)/(0.5/√35)

Z = 1.692

To estimate the probability we need to find the p value at Z > 1.692

P(Z > 1.692) = 0.0453

Therefore the probability is 0.045 or 4.5%