Given a normal distribution with mean = 100 and standard deviation = 10, if you select a sample of n = 25, what is the probability that x-bar is between 95 and 97.5?

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Answer 1 · 2017-09-19T11:05:24

$= 0.0994$ $=0.0994$

if $X ~ N (100, 10^{2})$ $X~N(100,10^2)$

then the sampling distribution of the mean with sample size $n$ $n$

$¯ ¯ ¯ X ~ N (100, \frac{10^{2}}{n})$ $barX~N(100,10^2/n)$

in this case

$¯ ¯ ¯ X ~ N (100, \frac{10^{2}}{25})$ $barX~N(100,10^2/25)$

$\Rightarrow ¯ ¯ ¯ X ~ N (100, 4)$ $=>barX~N(100,4)$

we want

$P (95 < ¯ ¯ ¯ X < 97.5)$ $P( 95 < barX<97.5)$

standardising

$P (\frac{95 - 100}{2} < Z < \frac{97.5 - 100}{2})$ $P(( 95-100)/2 < Z <(97.5-100)/2)$

where $Z ~ N (0, 1)$ $Z~N(0,1)$

$P (- 2.5 < Z < - 1.25)$ $P(-2.5 < Z <-1.25)$

by symmetry of the Normal distribution

$= P (2.5 < Z < 1.25)$ $=P( 2.5< Z<1.25)$

$= P (z < 2.5) - P (Z < 1.25)$ $=P(z<2.5)-P(Z<1.25)$

from tables

#=0.9938-0.8944

$= 0.0994$ $=0.0994$