City planners are planning to build a new community park. They agree that the park will be rectangular in shape, but the propose
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dimensions changed. At the most recent meeting, it was determined that the length would remain at 21.6 meters but that the width
would increase by 4.7 meters. So, the park would now cover an area of 714.96 square meters.
Which equation can be used to determine the original width of the park?
dimensions changed. At the most recent meeting, it was determined that the length would remain at 21.6 meters but that the width
would increase by 4.7 meters. So, the park would now cover an area of 714.96 square meters.
Which equation can be used to determine the original width of the park?
1 answer:
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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.