Question

The arithmetic mean and the standard deviation of a series of 20 items are 20cms and 5cms respectively.But while calculating them an items 13 was misread as 30 .

Answer 1

Explanation:

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Answer 2

Appropriate Question:

The arithmetic mean and the standard deviation of a series of 20 items are 20cms and 5cms respectively.But while calculating them an items 13 was misread as 30. Find the correct mean and correct standard deviation.

Answer:

$\sf \: \boxed{\begin{aligned}& \qquad \:\sf \:Correct \: mean \: = \: \dfrac{383}{20} = 19.15 \: \qquad \: \\ \\& \qquad \:\sf \: Correct \: standard \: deviation \: = \: 4.66 \: \end{aligned}} \qquad \:$

Explanation:

Given that,

Incorrect mean of 20 observations = 20 cm

Number of observations = 20

Correct value = 13

Incorrect value = 30

Incorrect standard deviation = 5 cm

Now,

$\sf \: Incorrect\:mean = \dfrac{Incorrect \: \displaystyle\sum_{i=1}^{20}\sf x_i}{20}$

$\sf \: 20 = \dfrac{Incorrect \: \displaystyle\sum_{i=1}^{20}\sf x_i}{20}$

$\implies\sf \: \sf \: Incorrect \: \displaystyle\sum_{i=1}^{20}\sf x_i = 400$

Now,

$\sf \: Correct \:\displaystyle\sum_{i=1}^{20}\sf x_i = Incorrect\:\displaystyle\sum_{i=1}^{20}\sf x_i + 13 - 30$

$\sf \: Correct \:\displaystyle\sum_{i=1}^{20}\sf x_i = 400 - 17$

$\implies\sf \: Correct \:\displaystyle\sum_{i=1}^{20}\sf x_i = 383$

Thus,

$\implies\sf \: Correct \: mean = \dfrac{Correct \:\displaystyle\sum_{i=1}^{20}\sf x_i}{20} = \dfrac{383}{20}$

Now,

$\sf \: Incorrect \: \sigma = \sqrt{\dfrac{Incorrect\:\displaystyle\sum_{i=1}^{20}\sf {x_i}^{2} }{20} - {(Incorrect\:mean)}^{2} }$

$\sf \: 5 = \sqrt{\dfrac{Incorrect\:\displaystyle\sum_{i=1}^{20}\sf {x_i}^{2} }{20} - {(20)}^{2} }$

On squaring both sides, we get

$\sf \: 25 = \dfrac{Incorrect\:\displaystyle\sum_{i=1}^{20}\sf {x_i}^{2} }{20} - 400$

$\sf \: 400 + 25 = \dfrac{Incorrect\:\displaystyle\sum_{i=1}^{20}\sf {x_i}^{2} }{20}$

$\sf \: 425 = \dfrac{Incorrect\:\displaystyle\sum_{i=1}^{20}\sf {x_i}^{2} }{20}$

$\implies\sf \: \sf \: Incorrect{\displaystyle\sum_{i=1}^{20}\sf {x_i}^{2} } = 8500$

So,

$\sf \:Correct{\displaystyle\sum_{i=1}^{20}\sf {x_i}^{2} } = 8500 + {(13)}^{2} - {(30)}^{2}$

$\sf \:Correct{\displaystyle\sum_{i=1}^{20}\sf {x_i}^{2} } = 8500 + 169 - 900$

$\implies\sf \: \sf \:Correct{\displaystyle\sum_{i=1}^{20}\sf {x_i}^{2} } = 7769$

Thus,

$\sf \: Correct \: \sigma = \sqrt{\dfrac{Correct\:\displaystyle\sum_{i=1}^{20}\sf {x_i}^{2} }{20} - {(Correct\:mean)}^{2} }$

$\sf \: Correct \: \sigma = \sqrt{\dfrac{7769}{20} - {\bigg(\dfrac{383}{20} \bigg) }^{2} }$

$\sf \: Correct \: \sigma = \sqrt{\dfrac{7769}{20} - \dfrac{146689}{400} }$

$\sf \: Correct \: \sigma = \sqrt{\dfrac{155380 - 146689}{400} }$

$\sf \: Correct \: \sigma = \sqrt{\dfrac{8691}{400} }$

$\sf \: Correct \: \sigma = \dfrac{ \sqrt{8691} }{20}$

$\sf \: Correct \: \sigma = \frac{ 93.22 }{20}$

$\implies\sf \: \sf \: Correct \: \sigma = 4.66$

Hence,

$\implies\sf \: \boxed{\sf \: \sf \: Correct \: standard \: deviation \:is \: 4.66 \: }$

Hence,

$\implies\sf \: \boxed{\begin{aligned}& \qquad \:\sf \:Correct \: mean \: = \: \dfrac{383}{20} = 19.15 \: \qquad \: \\ \\& \qquad \:\sf \: Correct \: standard \: deviation \: = \: 4.66 \: \end{aligned}} \qquad \:$