A random sample of adult female reaction times has a sample mean of x¯=394.3 milliseconds and sample standard deviation of s=84.6 milliseconds. Use the Empirical Rule to determine the approximate percentage of adult female reaction times that lie between 140.5 and 648.1 milliseconds.
Answer:
Given that ,
mean = \mu = 394.3
standard deviation = \sigma = 84.6
Using z table,
P(140.5< x <648.1 ) = P[(140.5 - 394.3) / 84.6< (x -\mu) / \sigma < (648.1 - 394.3) /84.6 )]
= P( -3< Z < 3)
= P(Z <3 ) - P(Z < -3)
using empirical rule
99.7%=\mu + / - 2 =99.7 / 2=49.85%
99.7%=\mu + / - 2 =99.7 / 2=49.85%
P(Z <3 ) + P(Z < -3)
=49.85% +49.85%
=99.7%
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