Rhombus J K L M is shown. The length of J K is 2 x + 4 and the length of J M is 3 x.
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Answer:
There is a 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.
Step-by-step explanation:
Problems of normally distributed samples 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.
In this problem, we have that:
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard deviation of 2.1-cm. This means that .
For shipment, 9 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.
By the Central Limit Theorem, since we are using the mean of the sample, we have to use the standard deviation of the sample in the Z formula. That is:
This probability is 1 subtracted by the pvalue of Z when .
has a pvalue of 0.9992. This means that there is a 1-0.9992 = 0.0008 = 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.