The estimate for the mean is of: $227.15.
The estimate for the standard deviation is of: $219.4.
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- 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
