B. Probability Distributions · Continuous Distributions and the Normal Distribution

Using the z-Table to Find Normal Probabilities

B2.8 B2.5About 75 minutes

Learning goals

I can read the standard normal table to find the area to the left of a z-score.
I can find areas to the right of z, between two z-scores and outside two z-scores, including negative z-scores.
I can solve normal probability problems by sketching, standardizing and using the table.
I can check my answers with the empirical rule and with technology.

Words to know

z-table (standard normal table)
A table of areas under the standard normal curve. The table in this lesson gives P(Z < z), the area to the left of z.

cumulative area
The area to the left of a value: P(Z < z) or P(X < x).

right tail
The area to the right of a value: P(Z > z) = 1 − P(Z < z).

symmetry rule
Because the standard normal curve is symmetric about 0, P(Z < −z) = P(Z > z) = 1 − P(Z < z).

Beyond 68–95–99.7

The empirical rule works only for whole numbers of standard deviations. What is the probability that a seedling is between 24 cm and 38 cm tall, when those values are −0.78 and 0.59 standard deviations from the mean? For that we need a table of areas under the standard normal curve.

In the last lesson you saw that every normal distribution turns into Z ~ N(0, 1) when you standardize. So one table is enough for all of them.

How the table works

The table below gives P(Z < z), the area under the standard normal curve to the left of z, for z from 0.00 to 3.49.

Round z to two decimal places.
Find the row for the units and first decimal (for 1.27, row 1.2).
Find the column for the second decimal (for 1.27, column .07).
Read the area: 0.8980. So P(Z < 1.27) = 0.8980.

For negative z, use symmetry: the area to the left of −z equals the area to the right of +z. So P(Z < −z) = 1 − P(Z < z). For example, P(Z < −0.45) = 1 − 0.6736 = 0.3264.

(Many textbooks print a second page of negative z-values instead; the answers are the same.)

z	.00	.01	.02	.03	.04	.05	.06	.07	.08	.09
0.0	0.5000	0.5040	0.5080	0.5120	0.5160	0.5199	0.5239	0.5279	0.5319	0.5359
0.1	0.5398	0.5438	0.5478	0.5517	0.5557	0.5596	0.5636	0.5675	0.5714	0.5753
0.2	0.5793	0.5832	0.5871	0.5910	0.5948	0.5987	0.6026	0.6064	0.6103	0.6141
0.3	0.6179	0.6217	0.6255	0.6293	0.6331	0.6368	0.6406	0.6443	0.6480	0.6517
0.4	0.6554	0.6591	0.6628	0.6664	0.6700	0.6736	0.6772	0.6808	0.6844	0.6879
0.5	0.6915	0.6950	0.6985	0.7019	0.7054	0.7088	0.7123	0.7157	0.7190	0.7224
0.6	0.7257	0.7291	0.7324	0.7357	0.7389	0.7422	0.7454	0.7486	0.7517	0.7549
0.7	0.7580	0.7611	0.7642	0.7673	0.7704	0.7734	0.7764	0.7794	0.7823	0.7852
0.8	0.7881	0.7910	0.7939	0.7967	0.7995	0.8023	0.8051	0.8078	0.8106	0.8133
0.9	0.8159	0.8186	0.8212	0.8238	0.8264	0.8289	0.8315	0.8340	0.8365	0.8389
1.0	0.8413	0.8438	0.8461	0.8485	0.8508	0.8531	0.8554	0.8577	0.8599	0.8621
1.1	0.8643	0.8665	0.8686	0.8708	0.8729	0.8749	0.8770	0.8790	0.8810	0.8830
1.2	0.8849	0.8869	0.8888	0.8907	0.8925	0.8944	0.8962	0.8980	0.8997	0.9015
1.3	0.9032	0.9049	0.9066	0.9082	0.9099	0.9115	0.9131	0.9147	0.9162	0.9177
1.4	0.9192	0.9207	0.9222	0.9236	0.9251	0.9265	0.9279	0.9292	0.9306	0.9319
1.5	0.9332	0.9345	0.9357	0.9370	0.9382	0.9394	0.9406	0.9418	0.9429	0.9441
1.6	0.9452	0.9463	0.9474	0.9484	0.9495	0.9505	0.9515	0.9525	0.9535	0.9545
1.7	0.9554	0.9564	0.9573	0.9582	0.9591	0.9599	0.9608	0.9616	0.9625	0.9633
1.8	0.9641	0.9649	0.9656	0.9664	0.9671	0.9678	0.9686	0.9693	0.9699	0.9706
1.9	0.9713	0.9719	0.9726	0.9732	0.9738	0.9744	0.9750	0.9756	0.9761	0.9767
2.0	0.9772	0.9778	0.9783	0.9788	0.9793	0.9798	0.9803	0.9808	0.9812	0.9817
2.1	0.9821	0.9826	0.9830	0.9834	0.9838	0.9842	0.9846	0.9850	0.9854	0.9857
2.2	0.9861	0.9864	0.9868	0.9871	0.9875	0.9878	0.9881	0.9884	0.9887	0.9890
2.3	0.9893	0.9896	0.9898	0.9901	0.9904	0.9906	0.9909	0.9911	0.9913	0.9916
2.4	0.9918	0.9920	0.9922	0.9925	0.9927	0.9929	0.9931	0.9932	0.9934	0.9936
2.5	0.9938	0.9940	0.9941	0.9943	0.9945	0.9946	0.9948	0.9949	0.9951	0.9952
2.6	0.9953	0.9955	0.9956	0.9957	0.9959	0.9960	0.9961	0.9962	0.9963	0.9964
2.7	0.9965	0.9966	0.9967	0.9968	0.9969	0.9970	0.9971	0.9972	0.9973	0.9974
2.8	0.9974	0.9975	0.9976	0.9977	0.9977	0.9978	0.9979	0.9979	0.9980	0.9981
2.9	0.9981	0.9982	0.9982	0.9983	0.9984	0.9984	0.9985	0.9985	0.9986	0.9986
3.0	0.9987	0.9987	0.9987	0.9988	0.9988	0.9989	0.9989	0.9989	0.9990	0.9990
3.1	0.9990	0.9991	0.9991	0.9991	0.9992	0.9992	0.9992	0.9992	0.9993	0.9993
3.2	0.9993	0.9993	0.9994	0.9994	0.9994	0.9994	0.9994	0.9995	0.9995	0.9995
3.3	0.9995	0.9995	0.9995	0.9996	0.9996	0.9996	0.9996	0.9996	0.9996	0.9997
3.4	0.9997	0.9997	0.9997	0.9997	0.9997	0.9997	0.9997	0.9997	0.9997	0.9998

Standard normal table. Each entry is the area to the left of z, that is P(Z < z), for z from 0.00 to 3.49. Find the row for the first decimal and the column for the second decimal. Example: z = 1.27 is row 1.2, column .07, which gives 0.8980. For a negative z use symmetry: P(Z < −z) = 1 − P(Z < z). Beyond z = 3.49 the area to the left is above 0.9998, so use about 1.0000.

Four kinds of questions

Always sketch the curve and shade the region first.

Left: P(Z < a) = table value.
Right: P(Z > a) = 1 − P(Z < a).
Between: P(a < Z < b) = P(Z < b) − P(Z < a).
Outside: P(Z < a or Z > b) = 1 − P(a < Z < b).

Since P(Z = a) = 0, < and ≤ give the same answer. Check: every probability must be between 0 and 1, and a shaded region bigger than half the curve must give more than 0.5.

Three common regions under the standard normal curve. Left areas come straight from the table; right areas are 1 minus the table value; areas between two z-scores are a difference of two table values.

Worked example

Example 1: Areas under the standard normal curve

Find (a) P(Z < 1.27), (b) P(Z > 1.27), (c) P(Z < −0.45), (d) P(−0.45 < Z < 1.27).

(a) Row 1.2, column .07: 0.8980.
(b) 1 − 0.8980 = 0.1020.
(c) P(Z < 0.45) = 0.6736, so P(Z < −0.45) = 1 − 0.6736 = 0.3264.
(d) 0.8980 − 0.3264 = 0.5716.

(a) 0.8980 (b) 0.1020 (c) 0.3264 (d) 0.5716

Worked example

Example 2: The seedling problem

Heights of 16-month-old maple seedlings are normally distributed with mean 32 cm and standard deviation 10.2 cm. What is the probability that a randomly selected seedling is between 24.0 cm and 38.0 cm tall?

Sketch N(32, 10.2²) and shade from 24 to 38. The region straddles the mean and is less than one SD wide on each side, so expect a bit less than 0.68.
Standardize: z₁ = (24.0 − 32)/10.2 ≈ −0.78; z₂ = (38.0 − 32)/10.2 ≈ 0.59.
Table: P(Z < 0.59) = 0.7224. P(Z < −0.78) = 1 − P(Z < 0.78) = 1 − 0.7823 = 0.2177.
Subtract: 0.7224 − 0.2177 = 0.5047.

P(24.0 < X < 38.0) ≈ 0.5047, so about 50.5% of seedlings. (Technology without rounding z gives about 0.5054; small differences come from rounding z to 2 decimals.)

Seedling heights N(32, 10.2²). The shaded area between 24 cm and 38 cm is about 0.5047.

Worked example

Example 3: Outside a tolerance

Maple syrup bottles are filled with volumes N(500, 4²) mL. A bottle is rejected if it holds less than 495 mL or more than 507 mL. What percent are rejected?

Low end: z = (495 − 500)/4 = −1.25. P(Z < −1.25) = 1 − 0.8944 = 0.1056.
High end: z = (507 − 500)/4 = 1.75. P(Z > 1.75) = 1 − 0.9599 = 0.0401.
Add the two tails: 0.1056 + 0.0401 = 0.1457.

About 14.57% of bottles are rejected.

Worked example

Example 4: Checking the empirical rule

Use the table to find P(−1 < Z < 1), P(−2 < Z < 2) and P(−3 < Z < 3).

P(Z < 1) = 0.8413 and P(Z < −1) = 0.1587: difference 0.6826.
P(Z < 2) = 0.9772 and P(Z < −2) = 0.0228: difference 0.9544.
P(Z < 3) = 0.9987 and P(Z < −3) = 0.0013: difference 0.9974.

0.6826, 0.9544, 0.9974: that is where 68%, 95% and 99.7% come from.

Technology check

Spreadsheets: =NORM.S.DIST(z, TRUE) gives P(Z < z), and =NORM.DIST(x, μ, σ, TRUE) gives P(X < x) without standardizing. Graphing calculators have a "normalcdf" command (lower bound, upper bound, μ, σ). Answers from technology can differ from table answers in the third or fourth decimal place because the table needs z rounded to two decimals. Either is acceptable if you show your method.

Just for fun

Statisticians say the standard normal curve is the most relaxed curve in math: its mean is zero, its standard deviation is one, and everything else is just a matter of looking it up.

Hands-on activity

Table versus reality

You need: your 40 stopwatch times from the lesson on the normal distribution (or 40 new ones), a calculator, the z-table above.

Find x̄ and s of your times.
Use the normal model N(x̄, s²) and the z-table to predict the proportion of times under 4.90 s, and the proportion between 4.90 s and 5.10 s.
Multiply each proportion by 40 to get predicted counts.
Count how many of your times actually fall in each range.

What to look for: actual counts within a few of the predictions mean the normal model describes your data well. Big differences suggest your times are skewed (for example if you often stopped late).

Your turn

Find P(Z < 2.33).

Find P(Z > 0.84).

Find P(Z < −1.5).

Find P(−1.96 < Z < 1.96).

Find P(−0.25 < Z < 0.5).

Icicle lengths are N(22, 6²) cm. Find the probability an icicle is longer than 30 cm.

Snow-shovelling times are N(35, 9²) min. Find P(25 < X < 40).

More practice: warm-up

Find P(Z < 1).

Find P(Z > 2).

Find P(Z < −0.5).

Find P(Z < 0).

Which table row and column give P(Z < 2.06)?

More practice: on your own

Heights of a group of adults are N(170, 8²) cm. Find the probability that a person is taller than 180 cm.

For N(170, 8²), find the probability a person is between 160 cm and 175 cm.

For N(170, 8²), about how many people in a group of 600 are shorter than 150 cm?

Battery lives are N(400, 25²) h. What proportion last less than 360 h or more than 440 h?

A commute takes N(42, 7²) minutes. What is the probability it takes more than 50 minutes? Should a student who must be at school in 50 minutes leave that early?

Challenge

Find P(|Z| > 2.5), that is, P(Z < −2.5 or Z > 2.5).

Three seedlings are chosen at random from N(32, 10.2²). Assuming independence, find the probability that all three are taller than 38 cm.

Explain why P(Z < −a) = 1 − P(Z < a) for any a, using a sketch.

Check your understanding

Quiz

What is P(Z < −1.2)?

What is P(Z > 0.3)?

What is P(−0.6 < Z < 1.1)?

Which expression gives P(a < Z < b)?

X ~ N(50, 5²). What is P(X > 61)?

Using the table, what is P(−1 < Z < 1)?

For parents and teachers

Big idea (B2.8): to find a normal probability: sketch and shade, standardize with z = (x − μ)/σ, look up P(Z < z) in the table, and combine (1 − for right tails, subtract for regions between). The full table is in this lesson; later lessons repeat just the values they use.

How to help: Before your teen looks anything up, ask them to estimate the answer from the sketch ("more or less than half?"). Then compare with a spreadsheet: =NORM.DIST(38, 32, 10.2, TRUE) − NORM.DIST(24, 32, 10.2, TRUE).

Common difficulties:

Reading the table value as the right-tail area. This table gives the left area.
Forgetting the symmetry step for negative z.
Adding when they should subtract (or the reverse). The sketch fixes this.
Rounding z wrongly: 0.588 becomes 0.59, not 0.58.