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According to the records of an electric company serving the Boston area, the mean electric consumption for all households during winter is 1650 kilowatt-hours per month. Assume that the monthly electric consumptions during winter by all households in this area have a normal distribution with a mean of 1650 kilowatt-hours and a standard deviation of kilowatt-hours. What percentage of 320 the households in this area have a monthly electric consumption of 1883 to 1975 kilowatt-hours

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Answer:

P(1833 < X < 1975) = 7.55%

Step-by-step explanation:

From the given information:

Let X be the random variable that obeys a normal distribution and which represents the monthly electric consumption during winter by all households in the Boston area.

X [tex]\sim[/tex] N ( μ = 1650 , σ² = 320² )

The probability that a monthly consumption of 1883 to 1975 kilowatt is given as:

[tex]P(1883 < X< 1975) = P( \dfrac{1883 -1650}{320} < Z< \dfrac{1975-1650}{320})[/tex]

[tex]P(1883 <X< 1975) = P( \dfrac{233}{320} < Z< \dfrac{325}{320})[/tex]

[tex]P(1883 < X<1975) = P( 0.728 < Z< 1.0156)[/tex]

P(1833 < X < 1975) = P(Z< 1.0156) - P(Z< 0.738)

P(1833 < X < 1975) = 0.8452 - 0.7697

P(1833 < X < 1975) = 0.0755

P(1833 < X < 1975) = 7.55%

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