A sign is posted in an elevator stating that the maximum number of people allowed is 25, and that the maximum weight capacity is 4000lbs. Suppose the distribution of the variable Weight (among the population using the elevator) is normal with mean 150lbs and standard deviation 30lbs. What is the probability that a randomly selected group of 25 people will exceed the maximum weight capacity? What should the maximum number of people allowed be if the probability of the maximum weight capacity being exceeded is to be at most .00001?
SD : sd / sqrt(n) = 30/5 = 6
mU : 150
avg (x) : 4000 / 25 = 160
p(x > 4000) = 1 - phi[(160 - 1/2 - 150) / 6]
Use calculator to calculate phi.
To solve this problem, we need to calculate the probability that the total weight of a randomly selected group of 25 people will exceed the maximum weight capacity of 4000lbs.
First, we need to find the distribution of the sum of the weights of 25 people. Since the variable "Weight" follows a normal distribution with a mean of 150lbs and a standard deviation of 30lbs, we can use the properties of the normal distribution to calculate the distribution of the sum.
The sum of 25 independent normally distributed variables (with the same mean and standard deviation) is itself normally distributed with a mean equal to the sum of the means and a standard deviation equal to the square root of the sum of the variances.
Mean of sum = 25 * 150lbs = 3750lbs
Standard deviation of sum = sqrt(25) * 30lbs = 150lbs
Next, we need to calculate the probability that the sum of the weights exceeds 4000lbs. We can standardize the variable using the formula z = (x - mean) / standard deviation, where z is the standardized value, x is the observed value, mean is the mean of the distribution, and standard deviation is the standard deviation of the distribution.
z = (4000 - 3750) / 150 = 250 / 150 = 1.67
Now, we can use a standard normal distribution table or a statistical software to find the probability associated with z = 1.67. Alternatively, we can use a calculator or Python programming to calculate the probability directly.
For z = 1.67, the probability is approximately 0.9525.
Therefore, the probability that a randomly selected group of 25 people will exceed the maximum weight capacity of 4000lbs is approximately 0.9525.
To find the maximum number of people allowed if the probability of the maximum weight capacity being exceeded is to be at most 0.00001, we need to find the z-score associated with this probability.
Using a standard normal distribution table or a statistical software, we can find the z-score for a probability of 0.00001. The z-score is approximately -4.753.
z = (x - mean) / standard deviation
-4.753 = (x - 3750) / 150
Solving for x, we get:
x = -4.753 * 150 + 3750 = 307.05
Therefore, the maximum number of people allowed should be rounded down to 307 to ensure that the probability of the maximum weight capacity being exceeded is at most 0.00001.