Question

The arithmetic mean and standard deviation of a series of 20 items were computed as 20 and 5 cm respectively.While claculating test,an item 13 cm was misread as 30 claculate mean and standard deviation when misread items is omitted

Answer 1

Answer:

$\sf \: \boxed{\begin{aligned}& \qquad \:\sf \:Correct \: mean \: = \: \dfrac{370}{19} = 19.47 \: \qquad \: \\ \\& \qquad \:\sf \: Correct \: standard \: deviation \: = \: 4.33 \: \end{aligned}} \qquad \:$

Step-by-step 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,

$\sf \: Correct \:\displaystyle\sum_{i=1}^{19}\sf x_i = Incorrect\:\displaystyle\sum_{i=1}^{20}\sf x_i - 30$

$\sf \: Correct \:\displaystyle\sum_{i=1}^{19}\sf x_i = 400 - 30$

$\implies\sf \: Correct \:\displaystyle\sum_{i=1}^{19}\sf x_i = 370$

Thus,

$\implies\sf \: Correct \: mean = \dfrac{Correct \:\displaystyle\sum_{i=1}^{19}\sf x_i}{19} = \dfrac{370}{19}$

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,

$\sf \:Correct{\displaystyle\sum_{i=1}^{19}\sf {x_i}^{2} } = 8500 - {(30)}^{2}$

$\sf \:Correct{\displaystyle\sum_{i=1}^{19}\sf {x_i}^{2} } = 8500 - 900$

$\implies\sf \: \sf \:Correct{\displaystyle\sum_{i=1}^{19}\sf {x_i}^{2} } = 7600$

Thus,

$\sf \: Correct \: \sigma = \sqrt{\dfrac{Correct\:\displaystyle\sum_{i=1}^{19}\sf {x_i}^{2} }{19} - {(Correct\:mean)}^{2} }$

$\sf \: Correct \: \sigma = \sqrt{\dfrac{7600}{19} - {\bigg(\dfrac{370}{19} \bigg) }^{2} }$

$\sf \: Correct \: \sigma = \sqrt{\dfrac{7600}{19} - \dfrac{136900}{361} }$

$\sf \: Correct \: \sigma = \sqrt{\dfrac{144400 - 136900}{361} }$

$\sf \: Correct \: \sigma = \sqrt{\dfrac{7500}{361} }$

$\sf \: Correct \: \sigma = \dfrac{ \sqrt{7500} }{19}$

$\sf \: Correct \: \sigma = \dfrac{86.6025 }{20}$

$\implies\sf \: \sf \: Correct \: \sigma = 4.33$

Hence,

$\implies\sf \: \boxed{\sf \: \sf \: Correct \: standard \: deviation \:is \: 4.33 \: }$

Hence,

$\implies\sf \: \boxed{\begin{aligned}& \qquad \:\sf \:Correct \: mean \: = \: \dfrac{370}{19} = 19.47 \: \qquad \: \\ \\& \qquad \:\sf \: Correct \: standard \: deviation \: = \: 4.33 \: \end{aligned}} \qquad \:$

Answer 2

Answer:

the corrected mean and S.D. are 19.15 cm and 4.66 cm respectively.

Step-by-step explanation:

I have given your answer in both the photos above. Click on both the photos respectively and see.

sorry mathdude but your Answer is Not entirely correct.

beacase the real value of the corrected mean and S.D. are 19.15 cm and 4.66 cm respectively.

Please make me a brainliest Answer.