Answer:
a) 0.5714 = 57.14% probability that it is overloaded because they have a mean weight greater than 158 lb
b) No, because the probability of being overloaded is considerably high(57.14%). Ideally, it should be under 5%, which would be considered an unusual event.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Assume that weights of males are normally distributed with a mean of 160 lb and a standard deviation of 35 lb.
This means that ![\mu = 160, \sigma = 35](https://tex.z-dn.net/?f=%5Cmu%20%3D%20160%2C%20%5Csigma%20%3D%2035)
Sample of 10:
This means that ![n = 10, s = \frac{35}{\sqrt{10}} = 11.07](https://tex.z-dn.net/?f=n%20%3D%2010%2C%20s%20%3D%20%5Cfrac%7B35%7D%7B%5Csqrt%7B10%7D%7D%20%3D%2011.07)
a) Find the probability that it is overloaded because they have a mean weight greater than 158 lb.
This is 1 subtracted by the pvalue of Z when X = 158. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{158 - 160}{11.07}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B158%20-%20160%7D%7B11.07%7D)
![Z = -0.18](https://tex.z-dn.net/?f=Z%20%3D%20-0.18)
has a pvalue of 0.4286
1 - 0.4286 = 0.5714
0.5714 = 57.14% probability that it is overloaded because they have a mean weight greater than 158 lb.
b.) Does this elevator appear to be safe?
No, because the probability of being overloaded is considerably high(57.14%). Ideally, it should be under 5%, which would be considered an unusual event.