Low Birth Weight The University of Maryland Medical Center considers “low birth weights” to be those that are less than 5.5 lb or 2495 g. Birth weights are normally distributed with a mean of 3152.0 g and a standard deviation of 693.4 g (based on Data Set 4 “Births” in Appendix B).

a. If a birth weight is randomly selected, what is the probability that it is a “low birth weight”?

b. Find the weights considered to be significantly low, using the criterion of a birth weight having a probability of 0.05 or less.

c. Compare the results from parts (a) and (b).

Short Answer

Expert verified

a.The probability of Low birth weight is 0.1711.

b. The birth weight having a probability of 0.05 or less is 2011.4 g weight is considered significantly low.

c.The birth weights between 2011.4g and 2496g are considered “low birth weights” but are not significantly low.

Step by step solution

01

Given information

Low birth weights are recognized as weights lesser than 5.5lb or 2495g.

The birth weights follow Normal distribution with a mean of 3152.0g and a standard deviation of 693.4g.

02

Describe the distribution of birth weights

a.

Let X be the random variable for birth weight which follows a normal distribution

X~Nμ,σ2~N3152,693.42

03

Calculation for z score

The probability of birth weight lesser than 2495g can be expressed as follows:

PX<2495

The z score associated with the birth weights 2495 computed as:

z=x-μσ=2495-3152.0693.4=-0.95

04

Calculate the probability of getting a low birth weight

The probability that weight is less than 2496gis expressed as:

PX<2495=PZ<-0.95...1)

From the standard normal table, the cumulative probability of z = -0.95 is obtained from the cell intersection of row value -0.9 and column value 0.05, which is 0.1711.

Thus,

PX<2496=0.1711

Thus, the probability of weight lesser than 2495 g is 0.1711.

05

Finding the weights considered to be significantly low

b.

Significantly low weights are all those birth weights that have the probability of lesser than 0.05 to appear.

That is,

PX<x0.05, where x is the maximum weight below which the weights are significantly low.

The probability of birth weight having a probability of 0.05 or less can be expressed as:

PX<x=PZ<z=0.05

From the standard normal table, the area of 0.05 is observed corresponding to the row value -1.6 and column value between 0.04 and 0.05, which implies the z score is -1.645.

Substitute the value of z as follows:

z=x-μσx=μ+σ×z=3152-1.645×693.4=2011.357

Therefore, all birth weights lesser than 2011.4 g are considered to be significantly low.

06

Comparing the results 

c.

The birth weights are significantly low if they are 2011.35 g or less, which have only a 5% chance to appear in the selection.

Also, birth weights lower than 2496 g are considered low birth rates with 17.11% chances to appear.

Thus, any weight that lies between 2011.5 g and 2496 g is recognized as low birth weight but actually is not significantly low (rare).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Standard Normal DistributionIn Exercises 17–36, assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, draw a graph, then find the probability of the given bone density test scores. If using technology instead of Table A-2, round answers

to four decimal places.

Between -2.75 and -0.75.

Standard Normal DistributionIn Exercises 17–36, assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, draw a graph, then find the probability of the given bone density test scores. If using technology instead of Table A-2, round answers

to four decimal places.

Greater than 0.18

Standard Normal DistributionIn Exercises 17–36, assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, draw a graph, then find the probability of the given bone density test scores. If using technology instead of Table A-2, round answers

to four decimal places.

Between 2.00 and 3.00.

In Exercises 11–14, use the population of {34, 36, 41, 51} of the amounts of caffeine (mg/12oz) in Coca-Cola Zero, Diet Pepsi, Dr Pepper, and Mellow Yello Zero.

Assume that  random samples of size n = 2 are selected with replacement.

Sampling Distribution of the Variance Repeat Exercise 11 using variances instead of means.

In Exercises 21–24, use these parameters (based on Data Set 1 “Body Data” in Appendix B):• Men’s heights are normally distributed with mean 68.6 in. and standard deviation 2.8 in.• Women’s heights are normally distributed with mean 63.7 in. and standard deviation 2.9 in.Mickey Mouse Disney World requires that people employed as a Mickey Mouse character must have a height between 56 in. and 62 in.

a. Find the percentage of men meeting the height requirement. What does the result suggest about the genders of the people who are employed as Mickey Mouse characters?

b. If the height requirements are changed to exclude the tallest 50% of men and the shortest 5% of men, what are the new height requirements?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free