Given a normal distribution with a mean of 120 and a standard deviation of 12, what percentage of values is within the interval 96 to 144

0.9545

Step-by-step explanation:

Given that :

Mean (m) = 120

Standard deviation (s) = 12

what percentage of values is within the interval 96 to 144

P(96 < x < 144)

P(x < 96)

obtain the standardized score (Z) :

Z = (x - m) / s

Z = (96 - 120) / 12 = - 2

p(Z < - 2) = 0.02275 ( Z probability calculator)

P(x < 144)

obtain the standardized score (Z) :

Z = (x - m) / s

Z = (144 - 120) / 12 = 2

p(Z < - 2) = 0.97725 ( Z probability calculator)

p(Z < - 2) - p(Z < 2) = 0.97725 - 0.02275 = 0.9545

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