Question

. What do you say about skewness of the distribution if Median is 49.21 while two quartiles are 37.15 and 61.27 respectively​

Answer 1

Answer:

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Step-by-step explanation:

$so \: here \: for \: a \: certain \: distributiom \\ Q1 = lower \: quartile = 37.15 \\ Q2 = median = 49.21 \\ Q3 = upper \: quartile = 61.27 \\ \\ so \: we \: know \: that \\ Bowley \: coefficient \: of \: skewness = \frac{Q3 +Q1 - 2. Q2}{Q3 - Q1} \\ \\ ie \: \: Skb = \frac{Q3 +Q1 - 2.Q2 }{Q3 -Q1} \\ \\ = \frac{(61.27 + 37.15) - (2 \times 49.21)}{61.27 - 37.15} \\ \\ = \frac{98.42 - 98.42}{24.12} \\ \\ = \frac{0}{24.12} \\ \\ = 0 \\ \\ so \: as \: Skb = 0 \\ we \: can \: say \: that \\ the \: distribution \: of \: data \: is \: symmetrical \: skewed \: ie \\ Mean=Mode=Median$