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

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 %

21 i watched a 3 hr vid explaining it

step-by-step explanation:

du what else would it be

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