Answer:
0.8029 = 80.29% probability that the thickness is between 3.0 and 7.0 mm.
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:
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2022/formulas/mathematics/college/bnaa16b36eg8ubb4w75g6u0qutzsb68wqa.png)
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.
Mean of 4.6 millimeters (mm) and a standard deviation of 1.5 mm.
This means that
![\mu = 4.6, \sigma = 1.5](https://img.qammunity.org/2022/formulas/mathematics/college/agc4e4r1jeeowd2bsetr3tot0kkkmsy6ur.png)
Find the probability that the thickness is between 3.0 and 7.0 mm.
This is the pvalue of Z when X = 7 subtracted by the pvalue of Z when X = 3. So
X = 7
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2022/formulas/mathematics/college/bnaa16b36eg8ubb4w75g6u0qutzsb68wqa.png)
![Z = (7 - 4.6)/(1.5)](https://img.qammunity.org/2022/formulas/mathematics/college/ssnl0zp5v3rc2x1eufp1vzm0fpudnwm5ds.png)
![Z = 1.6](https://img.qammunity.org/2022/formulas/mathematics/college/5doc8qlcjpi1mykku9xsmjl2f7ci0vivpl.png)
has a pvalue of 0.9452
X = 3
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2022/formulas/mathematics/college/bnaa16b36eg8ubb4w75g6u0qutzsb68wqa.png)
![Z = (3 - 4.6)/(1.5)](https://img.qammunity.org/2022/formulas/mathematics/college/5gpcxaxynadigdbl92vqrjgbc47jtcnvnv.png)
![Z = -1.07](https://img.qammunity.org/2022/formulas/mathematics/college/luj98bk7z51vxpe0zn0uzwjeww34x083mn.png)
has a pvalue of 0.1423
0.9452 - 0.1423 = 0.8029
0.8029 = 80.29% probability that the thickness is between 3.0 and 7.0 mm.