H < 6h + 3/5 Can you help me please?
1 answer:
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Answer:
There is a 98.26% probability that it is overloaded.
This elevator does not appear to be safe.
Step-by-step explanation:
Problems of normally distributed samples 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
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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.
In this problem, we have that:
Assume that weights of males are normally distributed with a mean of 170 lb, so
They have a standard deviation of standard deviation of 29 lb, so .
We have a sample of 8 adults, so we have to find the standard deviation of the sample to use in the place of in the Z score formula.
Find the probability that it is overloaded because they have a mean weight greater than 162 lb, so
has a pvalue 0.0174.
So, there is a 1-0.0174 = 0.9826 = 98.26% probability that it is overloaded.
Any probability that is above 95% is considerer unusually high. So, this elevator does not appear to be safe, since there is a 98.26% probability that it is overloaded.
I got D.
There's a few ways to solve it; I prefer using tables, but there are functions on a TI-84 that'll do it for you too. The logic here is, you have a standard normal distribution which means right away, the mean is 0 and the standard deviation is 1. This means you can use a Z table that helps you calculate the area beneath a normal curve for a range of values. Here, your two Z scores are -1.21 and .84. You might notice that this table doesn't account for negative values, but the cool thing about a normal distribution is that we can assume symmetry, so you can just look for 1.21 and call it good. The actual calculation here is:
1 - Z-score of 1.21 - Z-score .84 ... use the table or calculator
1 - .1131 - .2005 = .6864
Because this table calculates areas to the RIGHT of the mean, you have to play around with it a little to get the bit in the middle that your graph asks for. You subtract from 1 to make sure you're getting the area in the middle and not the area of the tails in this problem.
