Question

The average time a subscriber spends reading The Wall Street Journal is 49 minutes (The Wall Street Journal Subscriber Study). Assume the standard deviation is 16 minutes and that the times are normal are normally distribution. 1) What is the probability a subscriber will spend at least 1 hour reading the journal?

Answer 1

Answer:

$P(X>60)=P(\frac{X-\mu}{\sigma}>\frac{60-\mu}{\sigma})=P(Z>\frac{60-49}{16})=P(z>0.6875)$

And we can find the probability with the complement rule and the normal standard distirbution and we got:

$P(z>0.6875)=1-P(z<0.6875)= 1-0.7541= 0.2459$

Step-by-step explanation:

Let X the random variable that represent the time spent reading of a population, and for this case we know the distribution for X is given by:

$X \sim N(49,16)$  

Where $\mu=49$ and $\sigma=16$

We are interested on this probability

$P(X>60)$

And we can use the z score formula given by:

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

Using this formula we got:

$P(X>60)=P(\frac{X-\mu}{\sigma}>\frac{60-\mu}{\sigma})=P(Z>\frac{60-49}{16})=P(z>0.6875)$

And we can find the probability with the complement rule and the normal standard distirbution and we got:

$P(z>0.6875)=1-P(z<0.6875)= 1-0.7541= 0.2459$