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puteri [66]
3 years ago
6

Birth weights in the United States are normally distributed with a mean of 3420 g and a standard deviation of 495 g. The Newport

General Hospital requires special treatment for babies that are less than 2450 g (unusually light) or more than 4390 g (unusually heavy). What is the percentage of babies who do not require special treatment because they have birth weights between 2450 g and 4390 g
Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
6 0

Answer:

P(2450

And we can find this probability with this difference:

P(-1.96

If we use the normal standard table or excel we got:

P(-1.96

And that represent 95% of the data. so then the percentage below 2450 is 2.5% and above 4390 is 2.5 %

Step-by-step explanation:

Let X the random variable that represent the birth weights of a population, and for this case we know the distribution for X is given by:

X \sim N(3420,495)  

Where \mu=3420 and \sigma=495

We are interested on this probability

P(2450

And we can use the z score formula given byÑ

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(2450

And we can find this probability with this difference:

P(-1.96

If we use the normal standard table or excel we got:

P(-1.96

And that represent 95% of the data. so then the percentage below 2450 is 2.5% and above 4390 is 2.5 %

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