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2. The mean birth weight of a full term boy born in the US is 7.7 lbs, with standard deviation of 1.1 lbs. A. Find the probability that a boy born full term in the US weighs less than 7 lbs. B. Find the probability that the average weight of 25 baby boys born in a particular hospital have an average weight less than 7 lbs.

User Filaton
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Answer:

A

The value is
P( X < 7 ) =0.26226

B

The value is
P( X < 7 ) = 0.00073

Explanation:

From the question we are told that

The population mean is
\mu = 7.7 lbs

The standard deviation is
\sigma = 1.1 \ lbs

The sample size is n = 25

Generally the probability that a boy born full term in the US weighs less than 7 lbs is mathematically represented as


P( X < 7 ) = P( (X - \mu )/(\sigma) < (7 - 7.7 )/(1.1 ) )


(X -\mu)/(\sigma ) &nbsp;= &nbsp;Z (The &nbsp;\ standardized \ &nbsp;value\ &nbsp;of &nbsp;\ X )

=>
P( X < 7 ) = P(Z <- 0.6364 )

From the z table the area under the normal curve to the left corresponding to -0.6364 is


P(Z <- 0.6364 ) =0.26226

=>
P( X < 7 ) =0.26226

Generally the standard error of mean is mathematically represented as


\sigma_(x) = (\sigma )/(√(n) )

=>
\sigma_(x) = ( 1.1 )/(√(25) )

=>
\sigma_(x) = 0.22

Generally the probability that the average weight of 25 baby boys born in a particular hospital have an average weight less than 7 lbs is mathematically represented as


P( \= X < 7 ) = P( (\= X - \mu )/(\sigma_(x)) < (7 - 7.7 )/( 0.22 ) )

=>
P( X < 7 ) = P(Z <- 3.1818 )

From the z table the area under the normal curve to the left corresponding to -3.1818 is

=>
P(Z <- 3.1818 ) = 0.00073

=>
P( X < 7 ) = 0.00073

User Vas Giatilis
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