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6 June, 12:12

Scores on a university exam are Normally distributed with a mean of 78 and a standard deviation of 8. The professor teaching the class declares that a score of 62 or higher is required for a grade of at least a D. Using the 68-95-99.7 rule, what percentage of students score below 62

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  1. 6 June, 13:03
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    Porcentage of students score below 62 is close to 0,08%

    Step-by-step explanation:

    The rule

    68-95-99.7

    establishes:

    The intervals:

    [ μ₀ - 0,5σ, μ₀ + 0,5σ] contains 68.3 % of all the values of the population

    [ μ₀ - σ, μ₀ + σ] contains 95.4 % of all the values of the population

    [ μ₀ - 1,5σ, μ₀ + 1,5σ] contains 99.7 % of all the values of the population

    In our case such intervals become

    [ μ₀ - 0,5σ, μ₀ + 0,5σ] ⇒ [ 78 - (0,5) * 8, 78 + (0,5) * 8 ] ⇒[ 74, 82]

    [ μ₀ - σ, μ₀ + σ] ⇒ [ 78 - 8, 78 + 8 ] ⇒ [ 70, 86 ]

    [ μ₀ - 1,5σ, μ₀ + 1,5 σ] ⇒ [ 78 - 12, 78 + 12 ] ⇒ [ 66, 90 ]

    Therefore the last interval

    [ μ₀ - 1,5σ, μ₀ + 1,5 σ] ⇒ [ 66, 90 ]

    has as lower limit 66 and contains 99.7 % of population, according to that the porcentage of students score below 62 is very small, minor than 0,15 %

    100 - 99,7 = 0,3 %

    Only 0,3 % of population is out of μ₀ ± 1,5 σ, and by symmetry 0,3 / 2 = 0,15 % is below the lower limit, 62 is even far from 66 so we can estimate, that the porcentage of students score below 62 is under 0,08 %
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