Recently, a random sample of 13–18 year olds was asked, "How much do you currently have in savings?" The data in the table represent the responses to the survey Approximate the mean and standard deviation amount of savings.

First, we estimate the mean, which is the sums of the multiplications of the relative frequencies and the halfway point of each interval.
The size of the sample was of [tex]339 + 93 + 53 + 17 + 9 + 8 + 1 = 520[/tex]

The halfway points and relative frequencies are given as follows, and will be used to calculate both the mean and the standard deviation:

(0 + 199)/2 = 99.5, with a relative frequency of 339/520 = 0.6519.
(200 + 399)/2 = 299.5, with a relative frequency of 93/520 = 0.1788.
(400 + 599)/2 = 499.5, with a relative frequency of 53/520 = 0.1019.
(600 + 799)/2 = 699.5, with a relative frequency of 17/520 = 0.0327.
(800 + 999)/2 = 899.5, with a relative frequency of 9/520 = 0.0173.
(1000 + 1199)/2 = 1099.5, with a relative frequency of 8/520 = 0.0154.
(1200 + 1399)/2 = 1299.5, with a relative frequency of 1/520 = 0.0019.

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Thus, the mean is of:

[tex]M = 0.6519(99.5) + 0.1788(299.5) + 0.1019(499.5) + 0.0327(699.5) + 0.0173(899.5) + 0.0154(1099.5) + 0.0019(1299.5) = 227.15[/tex]

The mean of the amount of savings is of $227.15.

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The standard deviation is the square root of the sum of the difference squared between each observation and the mean, thus:

[tex]S = \sqrt{0.6519(99.5 - 227.15)^2 + 0.1788(299.5 - 227.15)^2 + 0.1019(499.5 - 227.15)^2 + ...} = 219.4[/tex]

The standard deviation is of $219.4.

A similar problem is given at https://brainly.com/question/24651197

Recently, a random sample of 13–18 year olds was asked, "How much do you currently have in savings?" The data in the table represent the responses to the survey Approximate the mean and standard deviation amount of savings.​