Question

The mean and standard deviation of a set of a 16 non-zero positive numbers in an observation are 26 and 3.5 respectively. The mean and standard deviation of another set of 24 non-zero positive numbers without changing the circumstances of both set of observations, are 29 and 3, respectively. The mean and standard deviation of their combined set of observations will respectively be ​

Answer 1

Answer:

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Step-by-step explanation:

$to \: find \: \\ combined \: standard \: deviation \: S.D \: (σc) \\ \\ given = \\ n1 = 16 \\ mean \: (x1) = 26 \\ S.D \: (σ1) = 3.5 \\ \\ n2 = 24 \\ mean \: (x2) = 29 \\ S.D \: (σ2) = 3$

$so \: using \\ combined \: mean \: (Xc) = \frac{n1.x1 \: + \: n2.x2}{n1 + n2} \\ \\ = \frac{(16 \times 26) + (24 \times 29)}{16 + 24} \\ \\ = \frac{416 + 696}{40} \\ \\ = \frac{1112}{40} \\ \\ Xc= 27.8$

$now \: applying \\ d1 = x1 - Xc \\ = 26 - 27.8 \\ d1= - 1.8 \\ \\ similarly \\ d2 = x2 - Xc \\ = 29 - 27.8 \\ d2 = 1.2$

$now \: we \: know \: that \\ combined \: standard \: deviation \\ σc = \sqrt{ \frac{n1(σ1 {}^{2} + d1 {}^{2} ) + n2(σ2 {}^{2} + d2 {}^{2} ) }{n1 + n2} } \\ \\ = \sqrt{ \frac{16((3.5) {}^{2} + ( - 1.8) {}^{2}) + 24((3) {}^{2} + (1.2) {}^{2} ) }{16 + 24} } \\ \\ = \sqrt{ \frac{16(12.25 + 3.24) +24(9 + 1.44) }{40} } \\ \\ = \sqrt{ \frac{(16 \times 15.49) + (24 \times 10.44)}{40} } \\ \\ = \sqrt{ \frac{247.84 + 250.56}{40} } \\ \\ = \sqrt{ \frac{498.4}{40} } \\ \\ = \sqrt{12.46} \\ = 3.5298 \\ = 3.52$

$hence \: σc \: ie \: combined \: standard \\ deviation \: is \: about \: 3.52$