An international data plan charges $10.00
1 answer:
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Where the variance is 0.32, the mean is 0.40, and there are provided sets of probabilities, the standard deviation will be 0.566.
<h3>What is standard deviation?</h3>The average degree of variability in your dataset is represented by the standard deviation. It reveals the average deviation of each statistic from the mean. In general, values with a high standard deviation are spread out from the mean, whereas those with a low standard deviation are grouped together close to the mean.
Here,
mean=0.4
variance=(Xn-mean)².P(x)n
=(0-0.4)²*(0.64)+(1-0.4)²*(0.32)+(2-0.4)²*0.04
variance=0.32
Standard deviation=√(variance)
=√0.32
=0.566
The standard deviation will be 0.566 where variance in 0.32 and mean is 0.40 and given sets of probability.
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Answer:
$14000
Step-by-step explanation:
Given parameters:
Amount earned by employee x = $5250
Time duration = 3weeks
Unknown:
Amount earned in 8 weeks = ?
Solution:
Let us first find the rate of employee x earning;
Rate =
Insert parameters;
Rate = = $1750/week
So,
In 8 weeks;
Amount earned = rate x time duration
Amount earned = $1750/week x 8weeks
Amount earned = $14000