Question

An instructor believes the distribution of scores on her final exam will have mean 62 and standard deviation 7. Fred is a student in the course that was selected at random. Provide answers to the following to two decimal places. (a) Find an upper bound on the probability that Fred will not score between 47 and 77. (b) Provide a lower bound on the probability that Fred will score between 45 and 79.

Answer 1

Answer:

a) $P(47<X<77)=P(-2.143<Z<2.143)=P(Z<2.143)-P(Z<-2.143)=0.984-0.0161=0.968$

And then the probability that Fred will not score between 47 and 77 is

1-0.968 = 0.0321

b) $P(45<X<79)=P(-2.429<Z<2.429)=P(Z<2.429)-P(Z<-2.429)=0.992-0.00757=0.985$

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

$X \sim N(62,7)$  

Where $\mu=62$ and $\sigma=7$

We are interested on this probability  on the probability that Fred will not score between 47 and 77. So we can begin finding this probability

$P(47<X<77)$

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:

$P(47<X<77)=P(\frac{47-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{77-\mu}{\sigma})=P(\frac{47-62}{7}<Z<\frac{77-62}{7})=P(-2.143<Z<2.143)$

And we can find this probability on this way:

$P(-2.143<z<2.143)=P(Z<2.143)-P(Z<-2.143)$

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

$P(-2.143<Z<2.143)=P(Z<2.143)-P(Z<-2.143)=0.984-0.0161=0.968$

And then the probability that Fred will not score between 47 and 77 is

1-0.968 = 0.0321

Part b

$P(45<X<79)=P(\frac{45-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{79-\mu}{\sigma})=P(\frac{45-62}{7}<Z<\frac{79-62}{7})=P(-2.429<Z<2.429)$

And we can find this probability on this way:

$P(-2.429<Z<2.429)=P(Z<2.429)-P(Z<-2.429)$

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

$P(-2.429<Z<2.429)=P(Z<2.429)-P(Z<-2.429)=0.992-0.00757=0.985$