What is the length of JN? Please help!
1 answer:
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Answer:
0.6568 = 65.68% probability that she would be correct.
Step-by-step explanation:
When the distribution is normal, we use 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.
45000, 45500, 44500, 46000 and 44000 units. This rate of sales is considered average for the product over the last twenty years or so.
This means that:
Standard deviation of 1500 units.
This means that
What is the likelihood that she would be correct if she predicts sales between 43000 and 46000?
This is the pvalue of Z when X = 46000 subtracted by the pvalue of Z when X = 43000. So
X = 46000
has a pvalue of 0.7486
X = 43000
has a pvalue of 0.0918
0.7486 - 0.0918 = 0.6568
0.6568 = 65.68% probability that she would be correct.