Respuesta :
Answer:
0.5 = 50% probability that a random sample of 100 independent persons will cause an overload
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.
For the sum of n values of a distribution, the mean is [tex]\mu \times n[/tex] and the standard deviation is [tex]\sigma\sqrt{n}[/tex]
An expert in ski lifts thinks that the weights of individuals using the lift have expected weight of 200 pounds and standard deviation of 30 pounds. 100 individuals.
This means that [tex]\mu = 200*100 = 20000, \sigma = 30\sqrt{100} = 300[/tex]
If the expert is right, what is the probability that a random sample of 100 independent persons will cause an overload
Total load of more than 20,000 pounds, which is 1 subtracted by the pvalue of Z when X = 20000. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20000 - 20000}{300}[/tex]
[tex]Z = 0[/tex]
[tex]Z = 0[/tex] has a pvalue of 0.5
1 - 0.5 = 0.5
0.5 = 50% probability that a random sample of 100 independent persons will cause an overload
