Question

find the bowely coefficient of skewness given Q1 =53.67,median =65, Q3 =76.33 .Also vomment upon the result​

Answer 1

Answer:

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Step-by-step explanation:

$to \: find = \\ bowleys \: coefficient \: of \: skewness \: \\ (Skb) \\ \\ given = \\ lower \: quartile \: (Q1) = 53.67 \\ \\ median \: (Q2) = 65 \\ \\ upper \: quartile \: (Q3) = 76.33 \\ \\ so \: we \: know \: that \\ \\ Skb = \frac{Q3 + Q1 - 2Q2}{Q3 -Q1 } \\ \\ = \frac{(76.33 + 53.67) - (2 \times 65)}{76.33 - 53.67} \\ \\ = \frac{130 - 130}{22.66} \\ \\ = \frac{0}{22.66} \\ \\ Skb = 0$

$so \: as \: here \\ Skb = 0 \\ \\ the \: distribution \: of \: data \: is \: \\ symmetrical$