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PIT_PIT [208]
3 years ago
5

Some sources report that the weights of​ full-term newborn babies in a certain town have a mean of 7 pounds and a standard devia

tion of 1.2 pounds and are normally distributed.a. What is the probability that one newborn baby will have a weight within 1.2 pounds of the meanlong dashthat ​is, between 5.8 and 8.2 ​pounds, or within one standard deviation of the​ mean?b. What is the probability that the average of nine ​babies' weights will be within 1.2 pounds of the​ mean; will be between 5.8 and 8.2 ​pounds?c. Explain the difference between​ (a) and​ (b).
Mathematics
1 answer:
marusya05 [52]3 years ago
3 0

Answer:

a) P(5.8

b) P(5.8

c) For part a we are just finding the probability that an individual baby would have a weight between 5.8 and 8.2. So we can't compare the result of part a with the result for part b.

For part b we are finding the probability that the mean of 9 babies (from random sampling) would be between 5.8 and 8.2, so on this case we have a distribution with a different deviation depending on the sample size. And for this reason we have different values

Step-by-step explanation:

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".  The letter \phi(b) is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: \phi(b)=P(z

Let X the random variable that represent the weights of​ full-term newborn babies of a population, and for this case we know the distribution for X is given by:

X \sim N(7,1.2)

a. What is the probability that one newborn baby will have a weight within 1.2 pounds of the meanlong dashthat ​is, between 5.8 and 8.2 ​pounds, or within one standard deviation of the​ mean?

We are interested on this probability

P(5.8

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

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

P(5.8

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

P(5.8

b. What is the probability that the average of nine ​babies' weights will be within 1.2 pounds of the​ mean; will be between 5.8 and 8.2 ​pounds?

And let \bar X represent the sample mean, the distribution for the sample mean is given by:

\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})

On this case  \bar X \sim N(7,\frac{1.2}{\sqrt{9}})

The z score on this case is given by this formula:

z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}

And if we replace the values that we have we got:

z_1=\frac{5.8-7}{\frac{1.2}{\sqrt{9}}}=-3

z_2=\frac{8.2-7}{\frac{1.2}{\sqrt{9}}}=3

For this case we can use a table or excel to find the probability required:

P(5.8

c. Explain the difference between​ (a) and​ (b).

For part a we are just finding the probability that an individual baby would have a weight between 5.8 and 8.2. So we can't compare the result of part a with the result for part b.

For part b we are finding the probability that the mean of 9 babies (from random sampling) would be between 5.8 and 8.2, so on this case we have a distribution with a different deviation depending on the sample size. And for this reason we have different values

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