Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
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 [tex]\mu = 160, \sigma = 35[/tex]
Sample of 10:
This means that [tex]n = 10, s = \frac{35}{\sqrt{10}} = 11.07[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{158 - 160}{11.07}[/tex]
[tex]Z = -0.18[/tex]
[tex]Z = -0.18[/tex] 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.
