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Height The average height of American females aged $18-24$ is normally distributed with mean $\mu=65.5$ inches and $\sigma=2.5$ inches.a. What percentage of females are taller than 68 inches?b. What is the probability a female is between $5^{\prime} 1^{\prime \prime}$ and $5^{\prime} 4^{\prime \prime}$ tall?
a. $P(X \geq 68) \approx 0.16 \rightarrow 16 \%$b. $P(61 \leq X \leq 64) \approx 0.23832$
Calculus 1 / AB
Calculus 2 / BC
Chapter 8
Techniques of Integration
Section 9
Probability
Integration
Integration Techniques
Harvey Mudd College
University of Michigan - Ann Arbor
Idaho State University
Lectures
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Hey, it's clear someone knew Marie here. So what I didn't really know is that we're given the mean, which is 65.5 in the standard deviation, which is 2.5. And I just plugged it into the density function for normal distribution off a random variable X with the mean and standard deviation so far apart, they were looking for a taller than 68 inches. So this is gonna be then to grow from 68 to infinity. We just plug in our equation, which is one over 2.5 square root of two pi. He'd the negative x minus 65.5 square over 12.5 d x, which gives us around 0.16 so it's 16% for Part B. We're looking at five foot one and five foot four. It gives us 61 inches and 64 inches, respectively. So it's from 61 2 64 So we have the same thing, but our intervals changes from 61 to 64. One over 2.5 square root of two pi peaks. The negative x minus 65.5 square over 12.5 t x, which is equal to 0.23832
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