Can I have a little help on how to do this?
1 answer:
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Answer:
0.039 = 3.9% probability that the random sample of 16 people in the elevator will exceed the weight limit
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 normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
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.
If n variables are added, the mean is and the standard deviation is
In this problem:
What is the probability that the random sample of 16 people in the elevator will exceed the weight limit?
This is 1 subtracted by the pvalue of Z when X = 2750. So
By the Central Limit Theorem
has a pvalue of 0.961
1 - 0.961 = 0.039
0.039 = 3.9% probability that the random sample of 16 people in the elevator will exceed the weight limit
Answer:
There is not sufficient evidence to support the claim that the technique performs differently than the traditional method.
Step-by-step explanation:
The null hypothesis is:
The alternate hypotesis is:
The test statistic is:
In which X is the sample mean, is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
A researcher used the technique with 260 students and observed that they had a mean of 94 hours. Assume the standard deviation is known to be 6.
This means, respectively, that
The test-statistic is:
The pvalue is:
2(P(Z < -2.69))
P(Z < -2.69) is the pvalue of Z when X = -2.69, which looking at the z-table, is 0.0036
2*(0.0036) = 0.0072
0.0072 < 0.01, which means that the null hypothesis is accepted, that is, there is not sufficient evidence to support the claim that the technique performs differently than the traditional method.