Question

In: Statistics and Probability

In 1952, Dr. Virginia Apgar suggested five criteria for measuring a baby’s health at birth” skin...

In 1952, Dr. Virginia Apgar suggested five criteria for measuring a baby’s health at birth” skin color, heart rate, muscle tone, breathing, and response when stimulated. She developed a 0-1-2 scale to rate a new-born on each of the five criteria. A baby’s Apgar score is the sum of the ratings on each of the five scales, which gives a whole-number value from 0 to 10. Apgar scores are still used today to evaluate the health of newborns.

What Apgar scores are typical? To find out, researchers recorded the Apgar scores of over 2 million newborn babies in a single year. Imagine selecting one of these newborns at random. Define the random variable X=Apgar score of a randomly selected baby one minute after birth. The table below gives the probability distribution of X

Value

0

1

2

3

4

5

6

7

8

9

10

Probability


Compute the mean of the random variable X and interpret this value in context of this problem.

d. Compute the standard deviation of the random variable X and interpret this value in context of this problem.

.001

.006

.007

.008

.012

.020

.038

.099

.319

.437

.053


Solutions

Expert Solution

(i)

x           p              xp               x2 p

0         0.001          0                   0

1         0.006         0.006           0.006

2        0.007          0.014            0.028

3        0.008          0.024              0.072

4        0.012           0.048            0.192

5          0.020          0.100           0.500

6          0.038         0.228          1.368

7         0.099           0.693         4.851

8         0.319           2.552        20.416

9          0.437          3.933        35.397

10        0.053         0.530          5.300

-----------------------------------------------------

Total                     8.128 68.130

Mean = E(X) = 8.128

Interpretation:

The average Apgar score of a randomly selected baby one inute after birth = 8.128

(ii)

Var(X) = E(X2) - (E(X))2

= 68.130 - 8.1282 = 2.0656

Standard deviation =

Interpretation:

The variation exhibited by the Agar scores of randomly selected baby one minute after birth is 1.4372.


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