MHB How is Probability Applied in Newspaper Reading Time Statistics?

[tcardwe3]

The time spent in minutes in reading newspapers for an adult per day can be approximated by a normal distribution with a mean of 15 minutes and a standard deviation of 3 minutes.
A-Find the probability that the reading time per day for a randomly selected adult is more than 18 minutes
B-If 200 adults are randomly selected, approximately how many of them will spend in reading newspaper per day between 12-19.5 minutes?
C-What is the shortest time spent in reading newspapers for an adult per day that would still place him in the top 10%?

 

[MarkFL]

With the given data measured in minutes, we are given:

$$\mu=15$$ and $$\sigma=3$$

Now, in order to use the actual data values, we need to standardize them, or convert them into $z$-scores.

a) Can you convert 18 to a $z$-score, and then use your table to find the area under the standard normal curve to the right of this $z$-score?

 

[tcardwe3]

I got 1 for the z score
18-15/3=1.0
and that's .8413 on my z score graph in my book

 

[MarkFL]

I got 1 for the z score
18-15/3=1.0
and that's .8413 on my z score graph in my book

That's the area to the left of $z=1$, so what is the area to the right?

 

[tcardwe3]

That's the area to the left of $z=1$, so what is the area to the right?

Ummm...I don't see it

 

[MarkFL]

Ummm...I don't see it

For any given $z$-score, what is the sum of the area to the left and the area to the right, that is, what is the total area under the standard normal curve?

 

[tcardwe3]

For any given $z$-score, what is the sum of the area to the left and the area to the right, that is, what is the total area under the standard normal curve?

Okay, so I got .8413 for A and 155 for B
Am I on the right track?

 

[MarkFL]

Okay, so I got .8413 for A and 155 for B
Am I on the right track?

Part A says, "Find the probability that the reading time per day for a randomly selected adult is more than 18 minutes."

You have found the area to the left of $z=1$ so you have found the probability that the reading time per day for a randomly selected adult is less than 18 minutes.

If the total area under the standard normal curve is 1, and the area to the left is 0.8413, then what is the area to the right? It may help to let the unknown area be $A$, and state what we know as an equation, and then solve for $A$:

$$0.8413+A=1$$

Now, solve for $A$...what do you get?

 

[tcardwe3]

Part A says, "Find the probability that the reading time per day for a randomly selected adult is more than 18 minutes."

You have found the area to the left of $z=1$ so you have found the probability that the reading time per day for a randomly selected adult is less than 18 minutes.

If the total area under the standard normal curve is 1, and the area to the left is 0.8413, then what is the area to the right? It may help to let the unknown area be $A$, and state what we know as an equation, and then solve for $A$:

$$0.8413+A=1$$

Now, solve for $A$...what do you get?

You get .1587

 

[MarkFL]

You get .1587

Correct good!

For B I get:

$$N=\frac{200}{3\sqrt{2\pi}}\int_{12}^{19.5} e^{-\frac{(x-15)^2}{18}}\,dx\approx155$$

This agrees with your result. (Yes)

Do you have any ideas for part C?