Let X have Normal distribution with mean 45 and variance 81. If a random sample of...

Let X have Normal distribution with mean 45 and variance 81. If a random sample of size 25 is taken, which of the following is the probability that the sample average is between 41.40 and 45.63?

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Expert Solution

Solution :

Given that ,

mean = = 45

varianc= 81 SO

standard deviation = = 9

n = 25

= 45

= / $\sqrt$ n= 9/ $\sqrt$ 25=1.8

P(41.40< <45.63 ) = P[(41.40 - 45) / 1.8< ( - ) / < (45.63-45) /1.8 )]

= P(-2 < Z <0.35 )

= P(Z < 0.35) - P(Z <-2 )

Using z table

=0.6368-0.0228

=0.6140

probability= 0.6140

orchestra answered 1 year ago

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