Need help due may 10
1 answer:
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Answer:
14.46% probability that the instructor can finish grading in 4 and a half hours
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.
For sums of n values from a distribution, the mean is and the standard deviation is
In this problem, we have that:
What is the probability that the instructor can finish grading in 4 and a half hours
Four and half hours is 4.5*60 = 270.
So this probability is the pvalue of Z when X = 270.
By the Central Limit Theorem
has a pvalue of 0.1446
14.46% probability that the instructor can finish grading in 4 and a half hours
Answer:
hello your question lacks the required options attached is a picture of the complete question
One would expect about 5% of tests to be significant just by chance if the null hypothesis is true, and for 60 tests, this is 0.05(60) = 3 tests. This could explain why these tests are statistically significant.
Step-by-step explanation:
What is misleading about this is the study's final report because this marketing study was carried out with a level of significance of 5% (0.05) which means that if we carry the same study with varying sample data we will mostly like arrive at a conclusion against our null conclusion 5% of the time and this is not good for the the study because it is a type 1 error and has to be eliminated and it cannot be eliminated totally .
hence One would expect about 5% of tests to be significant just by chance if the null hypothesis is true, and for 60 tests, this is 0.05(60) = 3 tests. This could explain why these tests are statistically significant.
