Answer:
a) 0.0985 = 9.85% probability that the thickness is less than 3.0 mm.
b) 0.0582 = 5.82% probability that the thickness is more than 7.0 mm.
c) 0.8433 = 84.33% probability that the thickness is between 3.0 mm and 7.0 mm.
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 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.
Normally distributed, with a mean of 4.8 millimeters (mm) and a standard deviation of 1.4 mm.
This means that
a) the thickness is less than 3.0 mm.
This is the pvalue of Z when X = 3. So
has a pvalue of 0.0985
0.0985 = 9.85% 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 0.9418
1 - 0.9418 = 0.0582
0.0582 = 5.82% probability that the thickness is more than 7.0 mm.
c) the thickness is between 3.0 mm and 7.0 mm.
This is the pvalue of Z when X = 7 subtracted by the pvalue of Z when X = 3. From questions a and b, we have these pvalues. So
0.9418 - 0.0985 = 0.8433
0.8433 = 84.33% probability that the thickness is between 3.0 mm and 7.0 mm.