Question

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.

Answer 1

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 )  =  Z (The  \ standardized \  value\  of  \ 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`