A normal distribution has a mean of and a standard deviation of . Find the probability that a value selected at random is in the following interval.

Question

A normal distribution has a mean of 101 and a standard deviation of 5. Find the probability that a value selected at random is in the following interval.

at least 96

Answer 1

mean 101

sd =5

P(>96) = 54%

(96-101)/5 = -1 = z score

by the empirical rule 68% is within 1 standard deviation from the mean

divide by 2

68/2 = 34%

add 50%

34+50 = 84% = a very close approximation

but use z tables or a z calculator for a more accurate percentage.

84.134% rounded off to 3 decimal places

Answer 2

The z value = (x - μ)/σ. The mean = 101, the std. deviation is 5, so z = (96 - 101)/5 = -5/5 = -1.

So the value of 96 is 1 std deviation to the left of the mean. The area under the normal curve to the left of 96 is therefore 0.1587 per z = -1.0 on the std. normal distribution table.

However, they're asking for values that are "at least" 96, so you're looking for the probability of a value greater than 96. That would be the area to the right of 96 under the curve, or 1 - 0.1587 = 0.8413 or 84.1%.

On the normal distribution curve, 1σ to the left of the mean contains 34.1% of the area, the rest of the curve to the right of the mean has 50.0%, hence the total 84.1%.

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