The shop sold 12,789 chocolate and 9,324 cookie dough cones. It sold 1,067 more peanut butter cones than cookie dough cones and
1 answer:
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Answer:
Mean =26.2
Median=27
Mode=32
Range=38
Step-by-step explanation:
From the stem and leaf plot, we obtain the original data as:
9,11,12,12,14,17,25,25,27,32,32,32,33,35,41,42,47
The median is middle observation of the array.
Thebefore the median is 27
The mode is the number that occurs most.
The mode is 32.
The range is the highest observation minus the least observation.
The range is 47-9=38
The mean is given by:
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.