Question

Let X be a random variable that represents the weights of newborns and suppose the distribution of weights of newborns is approximately normal with a mean of 7.44 lbs and a standard deviation of 1.33 lbs. Use Geogebra to help respond to the parts that follow. Don’t forget to press “enter” or somehow make Geogebra adjust to things you type. (a) (1 point) What’s the probability that a random newborn will weigh more than 9 lbs?
(b) (1 point) What’re the chances that a newborn weighs less than two standard deviations below the mean?
(c) (1 point) Eighty percent of all newborns weigh less than how many pounds? Hint: Sketch the bell curve with the appropriate shaded area, then get Geogebra to match your sketch and estimate the weight.

Answer 1

Let, X be a random variable that represents the weights of newborns.

X follows Normal distribution with mean = 7.44 lbs and standard deviation = = 1.33 lbs.

a)

We have to find probability that a random newborn will weigh more than 9 lbs.

I.e in statistical notation, we have to find P( x > 9 )

P( x > 9 ) = 1 - P( x < 9 )

Using Excel function , =NORMDIST( x , , , 1 )

P( x < 9 ) =NORMDIST(9,7.44,1.33,1) = 0.879589

So, P( x > 9 ) = 1 - 0.879589 = 0.1204

The  probability that a random newborn will weigh more than 9 lbs is 0.1204

b)

We have to find probability that newborn weighs less than two standard deviations below the mean.

i.e P( x < - 2 ) = P( x < 7.44 -2*1.33) = P( x < 4.78)

Using Excel function , =NORMDIST( x , , , 1 )

P( x < 4.78) =NORMDIST(4.78,7.44,1.33,1) =0.0228

The probability that newborn weighs less than two standard deviations below the mean is 0.0228

c)

Using Excel function, =NORMINV( probability, , )

x = NORMINV(0.8,7.44,1.33) =8.5594

Eighty percent of all newborns weigh less than 8.5594 pounds.