Determine ƒ(–a) where ƒ(x) = x2– 6x and simplify.
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For the bell-shaped graph of the normal distribution of weights of Hershey kisses, the area under the curve is 1, the value of the median and mode both is 4.5338 G and the value of variance is 0.0108.
In the given question,
The weight of the chocolate and Hershey Kisses are normally distributed with a mean of 4.5338 G and a standard deviation of 0.1039 G.
We have to find the answer of many question we solve the question one by one.
From the question;
Mean(μ) = 4.5338 G
Standard Deviation(σ) = 0.1039 G
(a) We have to find for the bell-shaped graph of the normal distribution of weights of Hershey kisses what is the area under the curve.
As we know that when the mean is 0 and a standard deviation is 1 then it is known as normal distribution.
So area under the bell shaped curve will be
= 1
This shows that that the total area of under the curve.
(b) We have to find the median.
In the normal distribution mean, median both are same. So the value of median equal to the value of mean.
As we know that the value of mean is 4.5338 G.
So the value of median is also 4.5338 G.
(c) We have to find the mode.
In the normal distribution mean, mode both are same. So the value of mode equal to the value of mean.
As we know that the value of mean is 4.5338 G.
So the value of mode is also 4.5338 G.
(d) we have to find the value of variance.
The value of variance is equal to the square of standard deviation.
So Variance =
Variance = 0.0108
Hence, the value of variance is 0.0108.
To learn more about normally distribution link is here
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