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प्रश्न
People’s monthly electric bills in Chennai are normally distributed with a mean of ₹ 225 and a standard deviation of ₹ 55. Those people spend a lot of time online. In a group of 500 customers, how many would we expect to have a bill that is ₹ 100 or less?
उत्तर
Let X be a normally distributed variable with a mean of ₹ 225 and a standard deviation of ₹ 55
Here µ = 225 and σ = 55
The standard normal variate z = `(x - mu)/sigma = (x - 225)/55`
P(a bill have ₹ 100 or less) = P(X ≤ 100)
When x = 100
z = `(100 - 25)/55`
= `(-125)/55`
= – 2.27
P(X ≤ 100) = P(Z < – 2.27)
P(z < – 2.27) = P(`-oo` < z < 0) – P(– 2.27 < z < 0)
= 0.5 – P(0 < z < 2.27)
= 0.5 – 04884
= 0.0116
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