The two-way frequency table below shows data on years working with the company and college degree status for Tom's coworkers. Co
mplete the following two-way table of row relative frequencies. (If necessary, round your answers to the nearest hundredth.)
1 answer:
Answer:
Lets start with the top row.
First, add the two values.
5+14=19
Now, divide each value by the total.
5/19=0.26315789473
Round the decimal to the nearest hundredth.
5/19=0.26
14/19=0.73684210526
Round it to the nearest hundredth.
14/19=0.74
Now, The second row.
Add the two values.
16+7=23
Divide the first value by the total.
16/23=0.69565217391
Round it to the nearest hundredth.
16/23=0.70
Divide the second value by the total.
7/23=0.30434782608
Round to the nearest hundredth.
7/23=0.30
Done!
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Answer:
The value is c = 21.1445.
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:
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.
The weight distribution of parcels sent in a certain manner is normal with mean value 12lb and standard deviation 3.5 lb.
This means that
What value of c is such that 99% of all parcels are at least 1 lb under the surcharge weight?
This 1 added to the value of X for the 99th percentile, which is X when Z has a p-value of 0.99, so X when Z = 2.327.
1 + 20.1445 = 21.1445
The value is c = 21.1445.