Number's Between 0 and 100
Algebra
Level
pending
X is a normally distributed random variable with mean 67 and standard deviation 2.
What is the probability that X is between 63 and 71? Write your answer as a decimal rounded to the nearest thousandth.
The answer is 0.95.
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1 solution
A random variable with a normal distribution is called normally distributed. For a normally distributed random variable X with mean π and standard deviation π:
P(πβπ<X<π+π) β 0.68
P(πβ2π<X<π+2π) β 0.95
P(πβ3π<X<π+3π) β 0.997
In other words: The probability that X is within one standard deviation of the mean is Pβ0.68. The probability that X is within two standard deviations of the mean is Pβ0.95. The probability that X is within three standard deviations of the mean is Pβ0.997.
The probability that X is between 63 and 71 can be written as P(63<X<71).
Write 63 and 71 as the mean, π=67, plus or minus a multiple k of the standard deviation, π=2: 63 = π+πk 63 = 67+2k-4 = 2k-2 = k
So, 63=πβ2π. By a similar calculation, 71=π+2π. This means P(63<X<71)=P(πβ2π<X<π+2π). Since X is normally distributed:
P(πβ2π<X<π+2π)β0.95 So, P(63<X<71)=P(πβ2π<X<π+2π)β0.95. Got it