From the data given below- find the no of items N - xy -120- r -0-5- standard deviation of Y -8- - x 2-90- where x and y are deviations from arithmetic mean-

Σ_Y=√(∑ y^2/N) 8=√(∑ y^2/N) By squaring both sides, we get 64=∑ y^2/N⇒∑ y^2=64N Now r=∑ xy/√(∑ x^2×∑ y^2)⇒ 0.5=120/√(90× 64N) Squaring both sides 0.25=14400/57 ...

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Standard XI

Mathematics

Standard Deviation about Median

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Question

From the data given below, find the no of items (N) :

$\sum$ xy=120, r = 0.5, standard deviation of Y = 8, $\sum x^{2} = 90,$ where x and y are deviations from arithmetic mean.

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Solution

$σ_{Y} = \sqrt{\frac{\sum y^{2}}{N}} 8 = \sqrt{\frac{\sum y^{2}}{N}}$

By squaring both sides, we get

$64 = \frac{\sum y^{2}}{N} \Rightarrow \sum y^{2} = 64 N Now r = \frac{\sum x y}{\sqrt{\sum x^{2} \times \sum y^{2}}} \Rightarrow 0.5 = \frac{120}{\sqrt{90 \times 64 N}}$

Squaring both sides

$0.25 = \frac{14400}{5760 N} \Rightarrow 1440 N = 14400 N = \frac{14400}{1440} = 10$

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From the data given below, find the no of items (N) : ∑xy=120, r = 0.5, standard deviation of Y = 8, ∑x2=90, where x and y are deviations from arithmetic mean.

σY=√∑y2N8=√∑y2N By squaring both sides, we get 64=∑y2N⇒∑y2=64NNow r=∑xy√∑x2×∑y2⇒0.5=120√90×64N Squaring both sides 0.25=144005760N⇒1440N=14400N=144001440=10

From the data given below, find the no of items (N) :

$\sum$ xy=120, r = 0.5, standard deviation of Y = 8, $\sum x^{2} = 90,$ where x and y are deviations from arithmetic mean.