How do I calculate $Pr (Z<1.2)$ for a standard normal random variable, using a $Z$-Table?

Question

I don’t know how to use the Z table, as my professor didn’t show us an image of one while describing how it worked and I had a hard time following her. I’ve looked it up and am still lost. Please explain using the easiest terminology possible.

Answer 1

Here is a brief orientation to the standard normal distribution. Suppose $Z \sim \mathsf{Norm}(0,1).$ Then roughly speaking $P(-1 < Z < 1) \approx .68$ and $P(-2 < Z < 2) \approx .95.$ (You may have heard of the 'Empirical Rule', which is based on these probabilities.) Also, the density function of $Z$ is symmetrical about $0$ with $P(Z < 0) = 1/2.$

The figure below shows this density function. Areas under the curve represent probabilities. From the above it follows that $P(Z < -2) = P(Z > 2) \approx (1 - .95)/2 = .025$ and $P(-1 < Z < 0) = P(0 < Z < 1) \approx .68/2 = .34.$ Thus, the six areas between vertical red bars, reading left to right, represent approximate probabilities $.025,\, .135,\, .34,\, .34,\, .135,$ and $.025,$ respectively.

From this figure you can see that $P(Z < 1.2)$ is between about $.68 + .135 +.025 = .84$ and about $1 - .025 = .975.$ In working problems with standard normal distributions, it is always a good idea to make a sketch imitating this figure, so you can guess the approximate answer. This can keep you from getting lost in the details and making gross errors.

To get the exact value of $P(Z < 1.2)$ you can use printed tables of the standard normal cumulative distribution function. These tables come in various styles, so I can't tell you exactly how to use the one in your book. But here is the general idea.

• Look in the left margin of your table to find the $z$-value $1.2$ and look at the first entry in that row (header $.00$). Depending on the style of your table, you may find the probability $.3849$ or the probability $.8849.$

• The answer is $P(Z < 1.2) = 0.8849$ (which is between $.68$ and $.975$ as predicted). If you see $.3849$ then that is for $P(0 < Z < 1.2) = 0.3849$ and you need to add $0.5000$ to get $P(Z < 1.2) = 0.8849.$ There should be a small figure at the top of the table that indicates whether the tabled values are for $P(Z < z),$ for $z > 0$ or for $P(0 < Z < z).$

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Notes: A few extensions of this problem may be worth mentioning:

(a) What if you had been asked to find $P(Z < 1.26)?$ Then the procedure is the same, except you look in the row for $1.2$ and the column headed $.06,$ to get $P(Z < 1.26) = 0.8962.$ Probabilities involving $z$-values with two decimal places will probably appear in another homework problem soon.

(b) Sometimes the $z$-values may be negative. Can you figure out how to use the printed table to get $P(Z < -1.26) = 0.1038?$

(c) What if you had been asked to find $c$ such that $P(Z < c) = 0.9?$ Then you look in the body of the table (where the probabilities are) for the closest number to $0.9$ or to $0.4$ (depending on the style of your table). Then look in the margins of the table to get the corresponding value of $c.$ You won't find the exact probability. About the best you can do using printed tables is to say that $c$ is between $1.28$ and $1.29.$ [The exact answer requires the use of a statistical calculator or software: $P(Z < 1.281552) = .9000.$]

Answer 2

Since P(Z<1.2) we use the second Z Table

Go along the Y axis till you find +1.2. No go along the X axis for the second decimal which is 0 here.

Hence you get the answer as 0.88493

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