Q4.
From the large data set, the daily total sunshine, s, in Leeming during June 1987 is record.
The data is coded using x = 10s + 1 and the following summary statistics are obtained.
n = 30

Find the mean and standard deviation of the daily total sunshine
ΣΧ =
x=947
S = 33 065.37
XX
Find the mean and standard deviation of the daily total sunshine.

Answers

Answer 1

The coding of the statistic is used to make it easier to work with the large sunshine data set

The mean of the sunshine is 3.05[tex]\overline 6[/tex]The standard deviation is approximately  18.184

Reason:

The given parameters are;

The sample size, n = 3.

∑x = 947

Sample corrected sum of squares, Sₓₓ = 33,065.37

The mean and standard deviation = Required

Solution:

[tex]Mean, \ \overline x = \dfrac{\sum x_i}{n}[/tex]

The mean of the daily total sunshine is therefore;

[tex]Mean, \ \overline x = \dfrac{947}{30} \approx 31.5 \overline 6[/tex]

[tex]s = \dfrac{x}{10 } - \dfrac{1}{10}[/tex]

[tex]E(s) = \dfrac{Ex}{10 } - \dfrac{1}{10}[/tex]

[tex]E(s) = \dfrac{31.5 \overline 6}{10 } - \dfrac{1}{10} = 3.05 \overline 6[/tex]

The mean ≈ 3.05[tex]\overline 6[/tex]

Alternatively

,[tex]The \ mean \ of \ the \ daily \ total \ sunshine, \, s = \dfrac{31.5 \overline 6 - 1}{10 } = 3.05\overline 6[/tex]

The mean of the daily total sunshine, [tex]\overline s[/tex] ≈3.05[tex]\overline 6[/tex]

[tex]Var(s) = Var \left(\dfrac{x}{10 } - \dfrac{1}{10} \right)[/tex]

[tex]Var(s) = \left(\dfrac{1}{10}\right)^2 \times Var \left(x \right)[/tex]

Therefore;

[tex]Var(s) = \left(\dfrac{1}{10}\right)^2 \times 33,065.37 = 330,6537[/tex]

Therefore;

[tex]s = \sqrt{330.6537} \approx 18.184[/tex]

The standard deviation, [tex]s_s[/tex] ≈ 18.184

Learn more about coding of statistic data here:

https://brainly.com/question/14837870


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