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I'm confused. I have a problem where I have to find the probability that x is below the z value 7.7. My z table only goes to z values of 3.4. How do I calculate this? These are the hints my teacher gave me...

A z-value of 7.7 means that we have a value that is 7.7 standard deviations away from the mean....you're not wrong here. Think about these questions: - Is this likely to happen? - What is the probability of having a value at less than 7.7 standard deviations away from the mean?

Thanks!

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What's the value at $z = 3.4$ in either direction? I'm betting it's like $0.9999$ and $0.0001$. So, extrapolate this behavior out farther. –  Arkamis Mar 4 '13 at 17:21
    
0.9998 is the highest probability on the z table, at z=3.49. So you're implying that, because 7.7 is so many standard deviations from the mean, that the probability is zero. Sweet! –  Becky Mar 4 '13 at 17:30
    
Be careful with wording. The probability that you're below the Z-value is one minus the probability you're above it. –  Arkamis Mar 4 '13 at 17:44
    
Ah, you are correct. I'm used to automatically talking about being below the Z-value because the z-tables we use in class give the probability to the left of the z-values. So, in better words: the P(z>7.7) is approximately zero and P(z<7.7) is almost 100%? –  Becky Mar 4 '13 at 17:51
    
Basically, yes. –  Arkamis Mar 4 '13 at 18:02
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3 Answers

The probability that a random variable $Z$ with standard normal distribution is less than $7.7$ is, for all practical purposes, equal to $1$. We have $$\Pr(Z \gt 7.7)\approx 6.8\times 10^{-15}.$$

The probability that we are $7.7$ or more standard deviations away from the mean (either direction allowed) is twice that. But twice utterly negligible is still utterly negligible.

A look at the graph of the graph of the characteristic "bell-shaped" density function of the standard normal shows that almost all the area is concentrated between $-3.5$ and $3.5$.

Remark: Suppose that you buy a single ticket in one of the mega-million lotteries this year, and again a single ticket next year. The probability that you will be the grand prize winner both times is greater than $\Pr(Z\gt 7.7)$.

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I see, so the chance is practically zero. I guess I've just been so bent on having a value that I was really confusing myself. Once again, thanks a ton Andre' for your help! Math is definitely not a strong subject for me :) –  Becky Mar 4 '13 at 17:28
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Almost every number that you use with a standard normal table is an approximation. The approximation $\Pr(Z\lt 7.7)\approx 1$ is a far better approximation than what you use when you look up $\Pr(Z\lt 1.96)$. –  André Nicolas Mar 4 '13 at 17:36
    
I see! So when you were helping me last night on my commute times problem (this 7.7 z value is from the same problem), it applies like this: a sample mean at or close to a 42-minute commute time (which gave me 7.7) is basically not going to occur (makes sense because the population mean is about 23, so 42 is waaaay up there), so the probability of the sample mean being below 42 minutes is approximately 1 (or there's nearly a 100% probability). Woo hoo! Lightbulbs are clicking! –  Becky Mar 4 '13 at 17:46
    
Yes, it is virtually equal to $1$. Also, note that the normal distribution is being used as a model, and will in fact fit the real data at best modestly well. So calculations to many places are virtually meaningless. After all, the normal distribution with your mean and standard deviation can be negative, albeit with negligible probability. Negative commute times? But a probability model does not have to be dead on to be useful. –  André Nicolas Mar 4 '13 at 17:55
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A slight overestimate of, and a reasonable approximation to, $P(Z > 7.7)$ is obtained by

  1. Calculating the value of the standard normal density function (yes, I did mean to write density function) at $7.7$. $$\frac{1}{\sqrt{2\pi}}e^{-7.7^2/2} \approx 5.32\times 10^{-14}$$

  2. Divide the result by $7.7$ $$\frac{5.32\times 10^{-14}}{7.7} \approx 6.909\times 10^{-15}$$

which can be compared to the more exact result of $6.8\times 10^{-15}$ given by André.

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Thanks! I think I got it! :) –  Becky Mar 26 '13 at 21:09
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The probability of a normally distributed random variable being within 7.7 standard deviations is practically 100%.

Remember these rules: 68.2% of the probability density is within one standard deviation; 95.5% within two deviations, and 99.7 within three deviations.

The reason that tables don't go to 7.7 is because deviations beyond around three are of little practical use. If you see a random variable being out by 7.7 deviations, this is so unlikely that you should suspect something is wrong with the experiment.

To calculate the exact answer, you have to simply figure out the area under the bell curve between 0 and 7.7 and then multiply by two. To do that, you need the cumulative density function: i.e., the integral of the probability density function. Unfortunately, that function does not exist in closed form.

Since you're expected to compute this for homework, your teacher must have given you some tools by which he or she expects you to calculate cumulative densities that are not covered by your table.

Here is a paper about approximating the cumulative density function.

Scientific calculators which provide support for statistical computing often have a function for this. You enter an argument like 7.7, and the function computes the cumulative density from $-\infty$ to the argument: i.e. the area under the curve to the left of the argument. If you have such a function, then simply get the value for 7.7, and then subtract from that the value for -7.7.

Your calculator most likely performs this calculation using numerical integration over the probability density function, rather than an approximation formula.

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Thanks so much! :) –  Becky Mar 26 '13 at 21:09
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