Question

The distribution of the number of viewers for the American Idol TV broadcasts follows the normal distribution of a mean of 29 million and a standard deviation of 5 million. What is the probability that the next week%u2019s show will:

a) Have between 30 and 34 million viewers?

b) Have at least 23 million viewers?



c) Exceed 40 million viewers?

Answer 1

Answer:

a) $P(30<X<34)=P(\frac{30-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{34-\mu}{\sigma})=P(\frac{30-29}{5}<Z<\frac{34-29}{5})=P(0.2<Z<1)$

And we can find this probability on this way:

$P(0.2<Z<1.0)=P(Z<1.0)-P(Z<0.2)$

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

$P(0.2<Z<1)=P(Z<1)-P(Z<0.2)=0.841-0.579=0.262$

b) $P(X\geq 23)=P(\frac{X-\mu}{\sigma}\geq \frac{23-\mu}{\sigma})=P(Z\geq \frac{23-29}{5})=P(Z\geq -1.2)$

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

$P(Z\geq -1.2)=1-P(Z<-1.2)=1-0.11507=0.885$

c) $P(X> 40)=P(\frac{X-\mu}{\sigma}> \frac{40-\mu}{\sigma})=P(Z\geq \frac{40-29}{5})=P(Z> 2.2)$

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

$P(Z>2.2)=1-P(Z<2.2)=1-0.986=0.0139$

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".  

2) Part a

Let X the random variable that represent the number of viewers for the American Idol TV of a population, and for this case we know the distribution for X is given by:

$X \sim N(29,5)$  

Where $\mu=29$ and $\sigma=5$ with the values on this case representing millions

We are interested on this probability :

$P(30<X<34)$

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:

$P(30<X<34)=P(\frac{30-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{34-\mu}{\sigma})=P(\frac{30-29}{5}<Z<\frac{34-29}{5})=P(0.2<Z<1)$

And we can find this probability on this way:

$P(0.2<Z<1.0)=P(Z<1.0)-P(Z<0.2)$

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

$P(0.2<Z<1)=P(Z<1)-P(Z<0.2)=0.841-0.579=0.262$

Part b

We are interested on this probability :

$P(X\geq 23)$

$P(X\geq 23)=P(\frac{X-\mu}{\sigma}\geq \frac{23-\mu}{\sigma})=P(Z\geq \frac{23-29}{5})=P(Z\geq -1.2)$

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

$P(Z\geq -1.2)=1-P(Z<-1.2)=1-0.11507=0.885$

Part c

We are interested on this probability :

$P(X> 40)$

$P(X> 40)=P(\frac{X-\mu}{\sigma}> \frac{40-\mu}{\sigma})=P(Z\geq \frac{40-29}{5})=P(Z> 2.2)$

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

$P(Z>2.2)=1-P(Z<2.2)=1-0.986=0.0139$