I have the following problem:
An office block has a lift that can carry a maximum permitted load of 500kg. It is known that the weights of men and women using the lift can be modeled by normal distribution. Then men's weights have mean 80kg and standard deviation 12kg. The women's weights have mean 56kg and standard deviation 6kg.
On one occasion, 3 men and 4 women enter the lift. Calculate the probability that
- Their total weight does not exceed the maximum permitted load
- The total weight of 3 men exceeds that of the 4 women.
This is my approach, can you guys help correct it?
- Regarding the three men, one man has mean 80kg and standard deviation of 12kg. Hence, we can calculate their mean as 80 $\times$ 3 = 240kg, and their standard deviation as $12\sqrt{3}$kg. Hence, we can calculate their mean as 56 $\times$ 4 = 224kg, and their standard deviation as $6\sqrt{4}$ or 12kg.
With the mean and standard deviation of the three men and four women, we can calculate the total weight as 240+224=464kg and standard deviation as $\sqrt{(12\sqrt{3})^2+12^2}$ or 24kg .
As the lift can be modeled by normal distribution, we can calculate the probability that the total weight will not exceed the permitted load by: $P(X<500)= P(X-500<0) = P(Z< \frac{500-464}{24}) = P(Z<1.5)$
According to the z-score table, the probability is 0.93319 or 93.32% - The probability of three men exceeding that of the 4 women, can be calculated $P(3X>4Y) = P(3X-4Y > 0) = P(Z>\frac{240-224}{24})$ = P(Z>0.67)
According to the z-score table, the probability is 1 - 0.74857 = 0.2513 or 25.13%
I am having a hard time fully grasping the concepts and theorems, so I feel like my solution has a lot of problems. I have not been able to find similar problems and materials, so I am seeking help here. I apologize for the inconvenience, and I thank you for any and all contribution.