Z-score above 3.9

[jamos7]

Hello,

I'm new and trying to learn all things probability and have stumbled upon a 'Normal Distribution' question that asks -

Find the probability that an item with a Mean = 4cm, and Standard Deviation of = 0.3cm is going to fit into a space that requires items to be a size inbetween 4cm and 6cm.

I've worked out so far that -

4 - 4 / .3 = 0

and

6 - 4 / .3 = 6.666666666666667

But to my knowledge Z-Scores only tend to go up to plus or minus 3.9. Therefore how does the 6.67 fit into everything?

An online calculator says that the probability would equal 0.50 (50%) to fit into the space but I can't work out how that comes about unless you use the '4 - 4 / .3 = 0' equation which equals 0.50 on he Z-Score table.

Do you void the 6.67 value as it doens't register on the Z-Score table?

Any information wouild be usefuil andmuch appreciated as I can't seem to find anything putting me in the right direction.

* Also attached an image of the online calculators answer (if correct)

 

[ksdhart2]

The first thing I note is that your equations are missing some very important grouping symbols:

\(\displaystyle 4 - 4 / 0.3 = 4 - \frac{4}{0.3} = -\frac{28}{3} \neq 0\)

\(\displaystyle {\color{red}\textbf{(}} 4 - 4 / 0.3 {\color{red}\textbf{)}} = \frac{4 - 4}{0.3} = 0\)

That aside, it's absolutely not true that "z-Scores only [...] go up to plus or minus 3.9." Most Z-tables cap out at these values, but there's nothing stopping a data point from having a z-score of \(6.\overline{6}\) or 15 or even 500. The reason that nearly every z-table stops at \(\pm 4\) is because the chance of a data point falling in that range is so miniscule as to be ignoreable for most purposes. However, some fields of the sciences do require such accuracy, and it is possible to calculate probabilities involving high z-scores.

For this problem, you've correctly found that a measurement of 6 cm is \(6.\overline{6}\) standard deviations above the mean. And as we just established, such scores are so exceedingly rare that you can reduce the question being asked to one of "What is the chance that the measurement is above the mean?", to which the answer is obviously 50%. The true probability can be found by evaluating an integral:

\(\displaystyle \int\limits_{4}^{6} \frac{1}{\sqrt{0.18\pi}} e^{-\frac{(z-4)^2}{0.18}} \: dz = \frac{1}{2} \biggr( erf(4.71404) - erf(0) \biggr) \approx 0.4999999999869155\)

As you can see, while this answer is not technically exactly 0.5, it's so close as to not be worth quibbling over.