The normal distribution is the famous "bell curve" — symmetric and centered at the mean, with most values near the center and fewer values far away. It describes many real-world measurements (heights, test scores, manufacturing tolerances). The Empirical Rule: about 68% of data falls within 1σ, 95% within 2σ, and 99.7% within 3σ of the mean. To find probabilities, convert a value to a z-score, then use the z-table which gives $P(Z < z)$ — the area to the left. For the area to the right, subtract from 1. For area between two z-values, subtract the smaller from the larger.
A dimension is normally distributed with mean 120 mm and standard deviation 5 mm. What value is at approximately the 95th percentile?
For the 95th percentile, $z\approx1.645$.
$$x=\mu+z\sigma=120+1.645(5)=128.23\text{ mm}$$
Final answer: 128.23 mm.
Left-Tail Probability
Exam scores are normally distributed with mean 75 and standard deviation 10. Find P(X < 60).
Convert to z-score, then read directly from the z-table (left-tail area).
$$z=\frac{60-75}{10}=-1.50$$
$$P(X<60)=P(Z<-1.50)=0.0668$$
Final answer: 0.0668 or about 6.7% of students scored below 60.
Right-Tail Probability
Wire diameters are normally distributed with mean 2.50 mm and standard deviation 0.04 mm. Find the probability a wire exceeds 2.56 mm.
The z-table gives left-tail area. For the right tail, subtract from 1.
$$z=\frac{2.56-2.50}{0.04}=1.50$$
$$P(X>2.56)=1-P(Z<1.50)=1-0.9332=0.0668$$
Final answer: 0.0668 or 6.68%.
Using Symmetry of the Normal Curve
For a standard normal distribution, find P(−1.2 < Z < 1.2).
The normal curve is symmetric. $P(Z<1.2)=0.8849$. By symmetry, $P(Z<-1.2)=1-0.8849=0.1151$.
$$P(-1.2<Z<1.2)=0.8849-0.1151=0.7698$$
Final answer: 0.7698 — about 77% of values fall within 1.2 standard deviations of the mean.
Finding the Mean from a Percentile
A normal distribution has standard deviation 6. The 90th percentile is 82. Find the mean.
At the 90th percentile, $z\approx1.282$. Rearrange: $\mu=x-z\sigma$.
$$\mu=82-1.282(6)=82-7.69=74.31$$
Final answer: mean ≈ 74.31.
Manufacturing Tolerance
Bolt lengths are normally distributed with mean 50 mm and standard deviation 1.5 mm. Bolts between 47 mm and 53 mm are acceptable. What fraction is acceptable?
Final answer: 95.44% of bolts are within tolerance (Empirical Rule: ~95% within ±2σ).
Probability Below the Mean
Heights of adult males are normally distributed with mean 170 cm and standard deviation 7 cm. What is P(X < 170)?
170 cm is exactly the mean, so $z=0$. The z-table gives $P(Z<0)=0.5000$ because the normal curve is perfectly symmetric — exactly half the area is below the mean.
$$P(X<170)=P(Z<0)=0.5000$$
Final answer: 0.5000 or 50%.
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Exam Generator Problems
Additional board-style practice items for this topic.
Question Bank: t2162
MSTE - Statistics and Probability / Probability / Besavilla CE Pre-Board Math & Surveying
A batch of 100 capacitors contains 73 which are within the required tolerance values, 17 which are below the required tolerance values and the remainder are above the required tolerance values. Determine the probability that when randomly selecting a capacitor and then a second capacitor both are within the required tolerance values when selecting with replacement.
0.2565
0.8785
0.4785
0.5329
0.745
With replacement, each draw has the same probability of selecting a capacitor within tolerance. $P(\text{within tolerance})=\frac{73}{100}=0.73$ $P(\text{both within tolerance})=(0.73)(0.73)$ $\boxed{0.5329}$
