Question

Recently, birth weights of Norwegians were reported to have a mean of 3668 grams and a standard deviation of 511 grams (g). Suppose that a Norweigan baby was chosen at random.
(a) Find the probability that the baby’s birth weight was less than 4000 g.
(b) Find the probability that the baby’s birth weight was greater than 3750 g.
(c) Find the probability that the baby’s birth weight was between 3000 g and 4000 g.
(d) Find the probability that the baby’s birth weight was less than 2650 g or greater than 4650 g.

Answer 1

Solution :

Given that,

mean = = 3668

standard deviation = = 511

a ) P( x < 4000)

P ( x - / ) < ( 4000- 3668 / 511 )

P ( z < 332 / 511 )

P ( z < 0.65 )

= 0.7422

Probability = 0.7422

b ) P (x >  3750 )

= 1 - P (x < 3750 )

= 1 - P ( x -  / ) < ( 3750 - 3668 / 511 )

= 1 - P ( z < 82 / 511 )

= 1 - P ( z < 0.16)

Using z table

= 1 - 0.5636

= 0.4364

Probability = 0.4364

c ) P (3000 < x < 4000 )

P ( 3000 - 3668 / 511 ) < ( x -  / ) < ( 4000 -3668 / 511 )

P ( - 668 / 511 < z < 332 / 511 )

P (-1.31 < z < 0.65 )

P ( z < 0.65 ) - P ( z < -1.31)

Using z table

= 0.7422 - 0.0951

= 0.6471

Probability = 0.6471

d ) P( x < 2650)

P ( x - / ) < ( 2650 - 3668 / 511 )

P ( z < -1018 / 511 )

P ( z < -1.99)

= 0.0233

P (x >4650)

= 1 - P (x < 4650 )

= 1 - P ( x -  / ) < ( 4650- 3668 / 511 )

= 1 - P ( z < 982 / 511 )

= 1 - P ( z < 1.92 )

Using z table

= 1 - 0.9726

= 0.0274

Probability = 0.0233 + 0.0274 =0.0507