What is the difference between relative frequency and conditional relative frequency
1 answer:
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Answer:
a) 0.9641.
b) 0.0082
c) 0.0277
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.
(a) ...to the left of 1.8.
p-value of Z = 1.8, which, looking at the z-table, is of 0.9641.
(b) ...to the right of 2.4.
1 subtracted by the p-value of Z = 2.4.
Looking at the z-table, Z = 2.4 has a p-value of 0.9918.
1 - 0.9918 = 0.0082, which is the answer.
(c) ...between 1.8 and 2.4.
p-value of Z = 2.4 subtracted by the p-value of Z = 1.8.
From itens a and b, we have both. So
0.9918 - 0.9641 = 0.0277