Central Limit Theorem

[stephanie953]

This problem is driving me insane. A recent study of the life span of portable radios found the average to be 2.9 years, with a standard deviation of 0.7 years. If a sample of 50 radios is taken, find the probability that the mean lifetime of these radios will be less than 2.5 years.


I set up the Z score formula. 2.5 - 2.9/0.7/square root of 50. I get -4.04. Is this correct? I don't think it is because the chart he gave us only goes up to z scores of 3.09. Can someone help me figure this out please? Thanks.

 

[galactus]

\(\displaystyle z=\frac{(x-{\mu})\sqrt{n}}{\sigma}\)

\(\displaystyle z=\frac{(2.5-2.9)\sqrt{50}}{.7}=-4.04\)

You are correct.

This means the probability is very low that it will be less than 2.5, considering the standard deviation is only .7

Looking beyond the scope of the table, the probability would be .0000267

or about .00267%

In other words, very low.

Do you have access to a calcualtor or computer that will do integration?.

Here is the formula used to find those z table values:

\(\displaystyle \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{z}e^{\frac{-t^{2}}{2}}dt\)

I used z=-4.04 and plugged it into my TI.

 

[stephanie953]

Thank you so much. He usually doesn't go outside the table in our book. I was going crazy trying to figure out what I was doing wrong.