Respuesta :
Answer:
a) The the approximate number of babies between 2363 and 3234 are:
n = 0.683*621= 423.95 and that's approximately 424 babies
b) The approximate number of babies between 1492 and 4976 are:
n = 0.955*621= 592.74 and that's approximately 593 babies
Step-by-step explanation:
Assuming this problem:"A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 621 babies born in New York. The mean weight was 3234 grams with a standard deviation of 871 grams. Assume that birth weight data are approximately bell-shaped.
Estimate the number of newborns who weighed between 2363 grams and 4105 grams. Round to the nearest whole number.
The number of newborns who weighed between 1492 grams and 4976 grams is . "
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Part a
Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(3234,871)[/tex]
Where [tex]\mu=3234[/tex] and [tex]\sigma=871[/tex]
We select a sample size of n = 621. And we want to find this probability:
[tex] P(2363 < X < 4105) [/tex]
[tex]P(2363<X<4105)=P(\frac{2363-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{4105-\mu}{\sigma})=P(\frac{2363-3234}{871}<Z<\frac{4105-3234}{871})=P(-1<z<1)[/tex]
And we can find this probability taking this difference:
[tex]P(-1<z<1)=P(z<1)-P(z<-1)=0.841-0.159=0.683 [/tex]
And the the approximate number of babies between 2363 and 3234 are:
n = 0.683*621= 423.95 and that's approximately 424 babies
Part b
[tex] P(1492 < X < 4976) [/tex]
[tex]P(1492<X<4976)=P(\frac{1492-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{4976-\mu}{\sigma})=P(\frac{1492-3234}{871}<Z<\frac{4976-3234}{871})=P(-1<z<1)[/tex]
And we can find this probability taking this difference:
[tex]P(-1<z<1)=P(z<1)-P(z<-1)=0.977-0.0228=0.955 [/tex]
And the the approximate number of babies between 1492 and 4976 are:
n = 0.955*621= 592.74 and that's approximately 593 babies
