The weights of male nurses in a hospital are known to be normally distributed with mean ƒÝ = 72 kg and standard deviation ƒã = 7.5 kg. The hospital has a lift (elevator) with a maximum recommended load of 450 kg. Six male nurses enter the lift. Calculate the probability p that their combined weight exceeds the maximum recommended load.
Well, isn't it funny that six male nurses are trying to test the limits of the hospital lift? It looks like they might be in for a weighty situation!
But hey, let's not clown around when it comes to calculations. To calculate the probability that their combined weight exceeds the maximum recommended load, we need to find the probability that the sum of their weights is greater than 450 kg.
Since the weights of the male nurses are normally distributed, we can use the properties of normal distribution to solve this. The sum of normally distributed variables is also normally distributed, with the mean being the sum of the individual means and the variance being the sum of the individual variances.
So, let's do the math. The mean of the sum of their weights would be 6 multiplied by the mean weight of a male nurse, which is 6 x 72 = 432 kg.
The variance of the sum would be 6 multiplied by the variance of the individual weights, which is 6 x (7.5^2) = 337.5 kg^2.
Now, to find the probability, we need to calculate the z-score for a sum of 450 kg using the mean and variance we calculated. The z-score formula is: z = (x - mean) / standard deviation.
In this case, z = (450 - 432) / sqrt(337.5) = 17 / sqrt(337.5).
Now, we need to find the probability of a z-score greater than this value. We can use a standard normal distribution table or a calculator to find this probability.
But wait, before we jump into that, let me tell you a little joke to lighten the mood.
Why don't scientists trust atoms?
Because they make up everything!
Alright, back to calculations. Using the standard normal distribution table or calculator, the probability of a z-score greater than the one we calculated is approximately 0.1419.
So, the probability p that the combined weight of the male nurses exceeds the maximum recommended load is approximately 0.1419.
To calculate the probability that the combined weight of six male nurses exceeds the maximum recommended load of 450 kg, we will use the concept of the sampling distribution of the sample mean.
The combined weight of the six male nurses can be considered as the sum of six independent and identically distributed normal random variables, each with a mean of 72 kg and a standard deviation of 7.5 kg.
The mean of the combined weight is given by the sum of the individual means, which is 6 * 72 = 432 kg.
The standard deviation of the combined weight is given by the square root of the sum of the individual variances (since the variables are independent), which is sqrt(6 * (7.5)^2) ≈ 18.52 kg.
To find the probability that the combined weight exceeds 450 kg, we need to calculate the z-score and then find the corresponding probability using standard normal tables or calculators.
The z-score is calculated as:
z = (x - 432) / 18.52
where x is the maximum recommended load, which is 450 kg.
Substituting the values, we get:
z = (450 - 432) / 18.52 ≈ 0.973
Now we need to find the probability corresponding to this z-score. We can use standard normal tables or calculators to find the probability. Assuming a two-tailed test, where we are interested in the probability of the combined weight being either significantly less or significantly greater than the maximum recommended load, we double the obtained probability.
Using standard normal tables or a calculator, the probability corresponding to a z-score of 0.973 is approximately 0.8332.
Therefore, the probability p that the combined weight of the six male nurses exceeds the maximum recommended load is approximately 2 * (1 - 0.8332) = 0.3336, or 33.36%.
To calculate the probability that the combined weight of six male nurses exceeds the maximum recommended load, we can use the concept of the sum of normally distributed variables.
First, let's calculate the distribution of the sum of the weights of the six male nurses. Since the weights of the male nurses are normally distributed with a mean of 72 kg and a standard deviation of 7.5 kg, the sum of these weights will also be normally distributed.
The sum of normally distributed variables follows the following properties:
- The sum of the means is equal to the mean of the sum.
- The sum of the variances is equal to the variance of the sum (assuming that the variables are independent).
The mean of the sum is simply the sum of the individual means: 72 kg * 6 = 432 kg.
The variance of the sum is the sum of the individual variances: (7.5 kg)^2 * 6 = 337.5 kg^2.
The standard deviation of the sum is the square root of the variance of the sum: √337.5 kg^2 ≈ 18.36 kg.
Now, we can calculate the probability that the combined weight exceeds the maximum recommended load of 450 kg. To do this, we will convert the problem into a standard normal distribution.
We can calculate the z-score using the following formula:
z = (x - μ) / σ
where x is the maximum recommended load, μ is the mean of the sum, and σ is the standard deviation of the sum.
z = (450 kg - 432 kg) / 18.36 kg ≈ 0.98
Next, we need to calculate the probability that the z-score is greater than 0.98. We can refer to a standard normal distribution table or use statistical software to find this probability.
Using a standard normal distribution table, the probability corresponding to a z-score of 0.98 is approximately 0.8365.
Therefore, the probability that the combined weight of the six male nurses exceeds the maximum recommended load is approximately 0.8365.
Z = (score-mean)/SEm
Score = 450/6
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.