Question

The Statistical Abstract of the United States published by the U.S. Census Bureau reports that the average annual consumption of fresh fruit per person is 99.9 pounds. The standard deviation of fresh fruit consumption is about 30 pounds. Suppose a researcher took a random sample of 38 people and had them keep a record of the fresh fruit they ate for one year.Round all z values to 2 decimal places. Round the answers to 4 decimal places.a. What is the probability that the sample average would be less than 90 pounds?p =The Statistical Abstract of the United States publb. What is the probability that the sample average would be between 98 and 105 pounds?p =The Statistical Abstract of the United States publc. What is the probability that the sample average would be less than 112 pounds?p =The Statistical Abstract of the United States publd. What is the probability that the sample average would be between 93 and 96 pounds?p =The Statistical Abstract of the United States publThe tolerance is +/- 0.0005.

Answer 1

Answer:

a) $P(\bar X<90)=P(Z<\frac{90-99.9}{\frac{30}{\sqrt{38}}})=P(Z<-2.03)= 0.0212$

b)$P(98<\bar X<105)=P(-0.39<Z<1.05)=P(Z<1.05)-P(Z<-0.39)=0.8531-0.3483=0.5049$

c) $P(\bar X<112)=P(Z<\frac{112-99.9}{\frac{30}{\sqrt{38}}})=P(Z<2.49)= 0.9936$

d) $P(93<\bar X<96)=P(-1.42<Z<-0.80)=P(Z<-0.80)-P(Z<-1.42)=0.2119-0.0778=0.1341$

Step-by-step explanation:

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 heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(99.9,30)$  

Where $\mu=99.9$ and $\sigma=30$

We select a random sample of n=36. And from the central limit theorem we know that the distribution for the sample is given by:

$\bar X \sim N(\mu =99.9 , \frac{\sigma}{\sqrt{n}}= \frac{30}{\sqrt{38}}=4.87)$

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

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

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

$P(\bar X<90)=P(\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{90-99.9}{\frac{30}{\sqrt{38}}})$

$=P(Z<\frac{90-99.9}{\frac{30}{\sqrt{38}}})=P(Z<-2.03)= 0.0212$

Part b

For this case we want this probability:

$P(98<\bar X<105)=P(\frac{98-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{105-\mu}{\frac{\sigma}{\sqrt{n}}})$

$=P(\frac{98-99.9}{\frac{30}{\sqrt{38}}}<Z<\frac{105-99.9}{\frac{30}{\sqrt{38}}})=P(-0.39<Z<1.05)$

And we can find this probability on this way:

$P(-0.39<Z<1.05)=P(Z<1.05)-P(Z<-0.39)$

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

$P(-0.39<Z<1.05)=P(Z<1.05)-P(Z<-0.39)=0.8531-0.3483=0.5049$

Part c

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

$P(\bar X<112)=P(\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{112-99.9}{\frac{30}{\sqrt{38}}})$

$P(\bar X<112)=P(Z<\frac{112-99.9}{\frac{30}{\sqrt{38}}})=P(Z<2.49)= 0.9936$

Part d

For this case we want this probability:

$P(93<\bar X<96)=P(\frac{93-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{96-\mu}{\frac{\sigma}{\sqrt{n}}})$

$=P(\frac{93-99.9}{\frac{30}{\sqrt{38}}}<Z<\frac{96-99.9}{\frac{30}{\sqrt{38}}})=P(-1.42<Z<-0.80)$

And we can find this probability on this way:

$P(-1.42<Z<-0.80)=P(Z<-0.80)-P(Z<-1.42)$

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

$P(-1.42<Z<-0.80)=P(Z<-0.80)-P(Z<-1.42)=0.2119-0.0778=0.1341$