Solve the proportion -- 33/6 = 55/z
1 answer:
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Answer:
0.6826 = 68.26% probability Z will be within one standard deviation of average.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
What is the probability Z will be within one standard deviation of average?
This is the p-value of Z = 1 subtracted by the p-value of Z = -1.
Z = 1 has a p-value of 0.8413.
Z = -1 has a p-value of 0.1587.
0.8413 - 0.1587 = 0.6826
0.6826 = 68.26% probability Z will be within one standard deviation of average.