Question

At-term newborns in Canada vary in weight according to, approximately, a Normal distribution, with a mean of 3500 grams and standard deviation of 500 grams

(a) Heavy birth weight (HBW) babies are those weighing over 4500 grams. Approxi- mately how many at-term newborns among the next 10000 will be HBW babies?

(b) Low birth weight (LBW) babies are those weighing less than 2500 grams. Ap- proximately how many at-term newborns among the next 10000 will be LBW babies?

(c) Approximately how many at-term newborns among the next 10000 will be babies weighing between 3300 and 4300 grams?

(d) A very low birth weight (VLBW) is sometimes defined as less than 1500 grams. If we wanted to set a VLBW limit at the 0.1th percentile of the Canadian distri- bution, i.e., at a weight such that only 1 in 1000 babies weigh less, what would this VLBW limit equal?

(e) One in ten at-term babies weighs more than

Answer 1

At-term newborns in Canada vary in weight according to, approximately, a Normal distribution, with a mean of 3500 grams and standard deviation of 500 grams

Let X be the weight of an randomly selected At-term newborns in Canada. X has a Normal distribution, with a mean and standard deviation

(a) Heavy birth weight (HBW) babies are those weighing over 4500 grams. Approximately how many at-term newborns among the next 10000 will be HBW babies?

The probability that a randomly selected baby weighs more than 4500 grams is

Let Y be the number of HBW babies among 10000 at-term newborns. We can say that Y has a binomial distribution with parameters, number of trials (number of at term newborns) n=10000 and success probability (The probability that a randomly selected baby weighs more than 4500 grams) p=0.0228

the expected value of Y is

ans: Approximately 228, at-term newborns among the next 10000, will be HBW babies

(b) Low birth weight (LBW) babies are those weighing less than 2500 grams. Ap- proximately how many at-term newborns among the next 10000 will be LBW babies?

The probability that a randomly selected baby weighs less than 2500 grams is

Let Y be the number of LBW babies among 10000 at-term newborns. We can say that Y has a binomial distribution with parameters, number of trials (number of at term newborns) n=10000 and success probability (The probability that a randomly selected baby weighs less than 2500 grams) p=0.0228

the expected value of Y is

ans: Approximately 228, at-term newborns among the next 10000, will be LBW babies

(c) Approximately how many at-term newborns among the next 10000 will be babies weighing between 3300 and 4300 grams?

The probability that a randomly selected at-term newborn weighs between 3300 and 4300 grams is

Let Y be the number of babies that weigh between 3300 and 4300 grams among 10000 at-term newborns. We can say that Y has a binomial distribution with parameters, number of trials (number of at term newborns) n=10000 and success probability (The probability that a randomly selected at-term newborn weighs between 3300 and 4300 grams) p=0.6006

the expected value of Y is

ans: Approximately 6006, at-term newborns among the next 10000 will be babies weighing between 3300 and 4300 grams

(d) A very low birth weight (VLBW) is sometimes defined as less than 1500 grams. If we wanted to set a VLBW limit at the 0.1th percentile of the Canadian distribution, i.e., at a weight such that only 1 in 1000 babies weigh less, what would this VLBW limit equal?

Let q grams be the VLBW limit such that, the probability that a randomly selected at-term new born weighs less than q is 0.001 (which is 0.1 percentile or 0.1/100=0.001)

We need

In terms of the z scores, we can write

However, we know that the area under the standard normal curve, to the left of mean (which is 0) is 0.5. Since the area required is 0.001 and it is less than 0.5, we can say that z should be negative (z<0)

Hence

Using the standard normal tables, we can get for z=3.09, P(Z<3.09)=0.999

or

We need

We can equate the z score of q to -3.09 and get

ans: The VLBW limit is equal to 1955 grams

(e) One in ten at-term babies weighs more than

Let One in ten at-term babies weighs more than q grams. This is same as the probability that a randomly selected baby weighs more than q grams is 1/10=0.1

In terms of the z scores, we need

Using the standard normal tables, we get for z=1.28, P(Z<1.28) =0.90

Or

But we need

We can equate the z score of X to 1.28 and get

ans: One in ten at-term babies weighs more than 4140 gram