Question

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

Answer 1

Answer:

a) $P(5.8<X<8.2)=P(\frac{5.8-7}{1.2}<Z<\frac{8.2-7}{1.2})=P(-1<Z<1)=P(Z<1)-P(Z<-1)=0.841-0.159=0.683$

b) $P(5.8<\bar X <8.2) = P(Z<3)-P(Z<-3)=0.999-0.0014=0.9973$

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: $\phi(b)=P(z<b)$

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

$P(5.8<X<8.2)$

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:

$P(5.8<X<8.2)=P(\frac{5.8-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{8.2-\mu}{\sigma})$

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

$P(5.8<X<8.2)=P(\frac{5.8-7}{1.2}<Z<\frac{8.2-7}{1.2})=P(-1<Z<1)=P(Z<1)-P(Z<-1)=0.841-0.159=0.683$

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:

$P(5.8<\bar X <8.2) = P(Z<3)-P(Z<-3)=0.999-0.0014=0.9973$

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