Complete the proof that ∆RSU~∆RTS.
1 answer:
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Answer:
a) 0.0869 = 8.69% probability that the thickness is less than 3.0 mm
b) 0.0668 = 6.68% probability that the thickness is more than 7.0 mm
Step-by-step explanation:
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.
In this problem, we have that:
(a) the thickness is less than 3.0 mm
This is the pvalue of Z when X = 3.
has a pvalue of 0.0869
0.0869 = 8.69% probability that the thickness is less than 3.0 mm
(b) the thickness is more than 7.0 mm
This is 1 subtracted by the pvalue of Z when X = 7. So
has a pvalue of 0.9332
1 - 0.9332 = 0.0668
0.0668 = 6.68% probability that the thickness is more than 7.0 mm