A teacher was interested in knowing the amount of physical activity that his stu

Question

A teacher was interested in knowing the amount of physical activity that his students were engaged in daily. He randomly sampled six students in his class and recorded the number of minutes that they exercised in the past day. The numbers were 30mins, 40 mins, 60 mins, 80 mins, 20 mins, and 85 mins. What’s the sample standard deviation? A teacher was interested in knowing the amount of physical activity that his students were engaged in daily. He randomly sampled six students in his class and recorded the number of minutes that they exercised in the past day. The numbers were 30mins, 40 mins, 60 mins, 80 mins, 20 mins, and 85 mins. What’s the sample standard deviation?

Explanation / Answer

Here Mean = (30+40+60+80+20+85)/6 = 52.5

Sum of square of deviation = (30-52.5)^2+(40-52.5)^2+(60-52.5)^2+(80-52.5)^2+(20-52.5)^2+(85-52.5)^2

Sum of square of deviation = 3587.5

Standard deviation = SQRT(3587.5/5) = 26.7862

So, answer is 26.7862

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