Answer:
1) P(258.5-cm < M < 259-cm) = 8.71%.
2) P17 = 125.57cm
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:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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.
Question 1:
Mean of 258.5-cm and a standard deviation of 2.3-cm, which means that ![\mu = 258.5, \sigma = 2.3](https://tex.z-dn.net/?f=%5Cmu%20%3D%20258.5%2C%20%5Csigma%20%3D%202.3)
Find the probability that the average length of a randomly selected bundle of steel rods is between 258.5-cm and 259-cm.
This is the pvalue of Z when X = 259 subtracted by the pvalue of Z when X = 258.5.
X = 259
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{259 - 258.5}{2.3}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B259%20-%20258.5%7D%7B2.3%7D)
![Z = 0.22](https://tex.z-dn.net/?f=Z%20%3D%200.22)
has a pvalue of 0.5871
X = 258.5
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{258.5 - 258.5}{2.3}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B258.5%20-%20258.5%7D%7B2.3%7D)
![Z = 0](https://tex.z-dn.net/?f=Z%20%3D%200)
has a pvalue of 0.5
0.5871 - 0.5 = 0.0871
0.0871*100% = 8.71%. So
P(258.5-cm < M < 259-cm) = 8.71%.
Question 2:
Mean of 127.1-cm and a standard deviation of 1.6-cm, which means that ![\mu = 127.1, \sigma = 1.6](https://tex.z-dn.net/?f=%5Cmu%20%3D%20127.1%2C%20%5Csigma%20%3D%201.6)
Find P17, which is the average length separating the smallest 17% bundles from the largest 83% bundles.
This is the 17th percentile, which is X when Z has a pvalue of 0.17. So X when Z = -0.954.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![-0.954 = \frac{X - 127.1}{1.6}](https://tex.z-dn.net/?f=-0.954%20%3D%20%5Cfrac%7BX%20-%20127.1%7D%7B1.6%7D)
![X - 127.1 = -0.954*1.6](https://tex.z-dn.net/?f=X%20-%20127.1%20%3D%20-0.954%2A1.6)
![X = 125.57](https://tex.z-dn.net/?f=X%20%3D%20125.57)
P17 = 125.57cm