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A short class with Sarah and her students. The other ways are under More.
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X has mean 50 and standard deviation 8. We want P(42 < X < 58). What is our first step?
Convert 42 and 58 into z-scores with z = (x − μ)/σ.
For 42 it's (42 − 50)/8 = −1, and for 58 it's (58 − 50)/8 = 1.
So the question becomes P(−1 < Z < 1). What do the z-scores tell us?
42 is one standard deviation below the mean and 58 is one above.
Right. Standardising lets one table work for every normal distribution.
The table says P(Z < 1) = 0.8413. What is P(Z < −1)?
The curve is symmetric, so it's 1 − 0.8413 = 0.1587.
How do we get the area between −1 and 1?
Take the area to the left of 1 and remove the area to the left of −1: 0.8413 − 0.1587 = 0.6826.
Does that seem reasonable?
Yes, it's about 68%, which is what the empirical rule says for one standard deviation either side.
Using the same X, find P(X > 58).
That's P(Z > 1) = 1 − 0.8413 = 0.1587.
And if we took 200 values of X, how many would you expect between 42 and 58?
0.6826 × 200 is about 136.5, so roughly 137.
One more: P(46 < X < 62).
The z-scores are −0.5 and 1.5, so 0.9332 − 0.3085 = 0.6247.
z = (x − μ)/σ turns 42 and 58 into −1 and 1 when μ = 50 and σ = 8.
P(a < Z < b) = P(Z < b) − P(Z < a).
P(Z < −1) = 1 − P(Z < 1) = 0.1587, and P(−a < Z < a) = 2P(Z < a) − 1.
about 68% of values lie within one standard deviation of the mean, matching 0.6826.
IB writes N(μ, σ²), so N(50, 64) has standard deviation 8.
multiply a probability by the group size, for example 0.6826 × 200 ≈ 137.
Minus the mean, divide by sigma, now the table ends the dilemma; left of the top, take left of the bottom, and the area between is what you've got.
Two fences stand at 42 and 58 in a field of sheep spread out around the farmhouse at 50. Each fence is one standard deviation from the house, and about 68 out of every 100 sheep graze between them.
"Sketch, Standardise, Subtract" - sketch the curve, convert to z, subtract the table areas.
S.Z.T.S. - Sketch, Z-scores, Table, Subtract.
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What is the z-score formula?
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What is P(Z < 1) from the standard normal table?
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What is P(−1 < Z < 1)?
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If X ~ N(50, 64) in IB notation, what is the standard deviation?
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What does the empirical rule say about one standard deviation?
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A continuous, symmetric, bell-shaped distribution described by its mean μ and standard deviation σ.
The normal distribution with mean 0 and standard deviation 1, written Z.
z = (x − μ)/σ, the number of standard deviations a value is above or below the mean.
Converting values of X into z-scores so that the standard normal table can be used.
A left-hand area P(Z < z), the value given by standard normal tables.
About 68%, 95% and 99.7% of a normal distribution lie within 1, 2 and 3 standard deviations of the mean.
IB notation for a normal distribution, in which the second number is the variance.
Example 1 finds P(42 < X < 58) for a normal variable X with mean 50 and standard deviation 8. Standardising gives z = −1 and z = 1, and the table gives 0.8413 − 0.1587 = 0.6826, so about 68% of values lie in this range.
To find a probability for any normal distribution, convert the values to z-scores with z = (x − μ)/σ and use the standard normal distribution. For X with mean 50 and standard deviation 8, the values 42 and 58 become z = −1 and z = 1. From the table, P(Z < 1) = 0.8413 and P(Z < −1) = 0.1587, so P(42 < X < 58) = 0.6826, about 68%, which matches the empirical rule for values within one standard deviation of the mean. This lesson works through each step with a sketch, shows a symmetry shortcut and the calculator method, explains the N(μ, σ²) notation, and practises related probabilities: one-sided areas, uneven intervals and expected numbers in a group.
The question. X is normally distributed with mean μ = 50 and standard deviation σ = 8. Find P(42 < X < 58).
A note on notation. This chapter writes the distribution as X ~ N(50, 8), giving the standard deviation as the second number. In IB notation the second number is the variance, X ~ N(μ, σ²), so the same distribution is written X ~ N(50, 8²) = N(50, 64). Always check whether the second number is σ or σ².
Step 1: sketch. Draw a bell curve centred at 50 and shade the region between 42 and 58. The region is symmetric about the mean and covers the middle of the curve, so expect a fairly large probability.
Step 2: standardise. Use z = (x − μ)/σ:
Step 3: read the table. The standard normal table gives P(Z < 1) = 0.8413. By symmetry, P(Z < −1) = 1 − 0.8413 = 0.1587.
Step 4: subtract. P(−1 < Z < 1) = P(Z < 1) − P(Z < −1) = 0.8413 − 0.1587 = 0.6826.
Step 5: interpret. About 68.3% of the values of X lie between 42 and 58. A calculator's normal cumulative distribution function, with lower bound 42, upper bound 58, μ = 50 and σ = 8, gives 0.6827 (the small difference comes from rounding in the table).
Why the answer is about 68%. 42 and 58 are exactly one standard deviation below and above the mean. The empirical rule says about 68% of a normal distribution lies within one standard deviation of the mean, about 95% within two, and about 99.7% within three.
A symmetry shortcut. For an interval symmetric about the mean, P(−a < Z < a) = 2P(Z < a) − 1. Here 2 × 0.8413 − 1 = 0.6826.
Related probabilities for the same X.
For a continuous distribution the probability of any single exact value is 0, so P(42 < X < 58) and P(42 ≤ X ≤ 58) are the same.