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Practice Problems: Area under the Normal Curve

Practice Problems

For these problems use the normal tables on the Documents Page. There are two tables, one for negative z-values and one for positive z-values.
In the answers below, the phrase "area of [a,b]" is short for "area under the normal curve for the bin [a,b]."

1. For a normal histogram with center 0 and spread 1, what proportion of observations are in the specified bin?
   a. (-∞, -2.00] Ans: 0.0288   b. (-∞, -3.00] Ans: 0.0013   c. (-∞, 0.00] Ans: 0.5000
   d. (-∞, 1.00] Ans: 0.8413   e. (-∞, 3.00] Ans: 0.9987   f. (-∞, 1.64] Ans: 0.9495
   g. (-∞, -0.72] Ans: 0.2358   h. (-∞, ∞) Ans: 1.000
   i. [2.87, ∞) = 1 - area of (-∞, 2.87] = 1 - 0.9972 = 0.0027
   j. [0.78, ∞) = 1 - area of (-∞, -0.78] = 1 - 0.2177 = 0.7823
   k. [-1.12, ∞) = 1 - area of (-∞, -1.12] = 1 - 0.1214 = 0.8786
   l. [-0.04, ∞) = 1 - area of (-∞, -0.04] = 1 - 0.4880 = 0.5120
   m. [-1.00, 1.00] = area of (-∞, 1.00] - area of (-∞, -1.00] = 0.8413 - 0.1587 = 0.6826
   n. [-2.00, 2.00] = area of (-∞, 2.00] - area of (-∞, -2.00] = 0.9772 - 0.0228 = 0.9544
   o. [-3.00, 3.00] = area of (-∞, 3.00] - area of (-∞, -3.00] = 0.9987 - 0.0013 = 0.9974
   p. [-3.49, 3.49] = area of (-∞, 3.49] - area of (-∞, -3.49] = 0.9998 - 0.0002 = 0.9996
   q. [-1.54, 2.45] = area of (-∞, 2.45] - area of (-∞, -1.54] = 0.9946 - 0.0618 = 0.9828
   r. [0.23, 2.09] = area of (-∞, 2.09] - area of (-∞, 0.23] = 0.9817 - 0.5910 = 0.3907
   s. [-0.95, 1.37] = area of (-∞, 1.37] - area of (-∞, -0.95] = 0.9147 - 0.1711 = 0.7436
   t. [-3.32, -3.09] = area of (-∞, -3.09] - area of (-∞, -3.32] = 0.0010 - 0.0005 = 0.0005
2. For intervals of the form (-∞, z], which values of z give the following proportion of observations in that interval?
   a. 0.138   Ans: (-∞, -1.09]  b. 0.376   Ans: (-∞, -0.31]  c. 0.500   Ans: (-∞, 0.00]
   d. 0.652   Ans: (-∞, 0.39]  e. 0.998   Ans: (-∞, 2.88]
3. For intervals of the form [z, ∞), which values of z give the the following proportions?
   a. 0.007   Ans: area of [-∞, 2.46] = 1 - 0.007 = 0.993, so area of [2.46, ∞) = 1 - 0.993 = 0.007.
   b. 0.287   Ans: area of [-∞, 0.56] = 1 - 0.287 = 0.713, so area of [0.56, ∞) = 1 - 0.713 = 0.287.
   c. 0.701   Ans: area of [-∞, -1.97] = 1 - 0.701 = 0.309, so area of [-1.97, ∞) = 1 - 0.309 = 0.701.
   d. 0.998   Ans: area of [-∞, -3.49] = 1 - 0.998 = 0.002, so area of [-3.49, ∞) = 1 - 0.002 = 0.998.
4. For bins of the form [-z, z], which values of z give the the following proportions?
   a. 0.137
   Ans: If area of [-z, z] = 0.137, then area of (-∞, -z] and [z, -∞) together is 1 - 0.0137 = 0.9863. Also area of (-∞, -z] = area of [z, ∞), so area of (-∞, -z] = 0.9863 / 2 = 0.4931. Therefore z = 0.02.
    
   b. 0.379
   Ans: By the same reasoning as Part a, if the area of [-z, z] is 0.379, the area of (-∞, -z] = (1 - 0.379) / 2 = 0.3105, so z = 0.52.
    
   c. 0.652
   Ans: By the same reasoning as Part a, if the area of [-z, z] is 0.652, the area of (-∞, -z] = (1 - 0.652) / 2 = 0.174, so z = 0.92.
    
   d. 0.992
   Ans: By the same reasoning as Part a, if the area of [-z, z] is 0.992, the area of (-∞, -z] = (1 - 0.992) / 2 = 0.004, so z = 3.38
5. For a population of giraffes, the heights are measured. The sample mean is 16.5 ft; the sample standard deviation is 3.8. What percentage of giraffes are taller than 20 ft.
   Ans: μ = 16.5, σ = 3.8, x = 20. Then z = (x - μ) / σ = (20 - 16.5) / 3.8 = 0.92. Look up 0.92 in the standard normal table: 0.8212 = 82%.
6. What is the 93rd percentile of height for the giraffes in Practice Problem 2?
   Ans: Look up 0.93 in the body of the standard normal table: z = 1.48. Then solve for x = z * σ + μ = 1.48 * 3.8 + 16.5 = 22.1 feet.
7. For a standard normal curve, where are Q1 and Q3 located?
   Ans: We want the area of [-z, z] to be 0.5000, so the area of (-∞, -z] = (1 - 0.5000) / 2 = 0.25. This means that z = 0.67; Q1 = -0.67, Q3 = 0.67. IQR = 0.67 - (-0.67) = 1.34
8. For a standard normal curve, how far from the center must an observation be to be called a mild outlier?
   Ans: The inner fence above Q3 is at Q3 + 1.5 * IQR = 0.67 + 2.01 = 2.68. The inner fence below Q1 is at -Q1 - 1.5 * IQR = -0.67 - 1.5 * 2.68 = -2.68.
9. Use the result of Problem 5 to predict the percentage of mild outliers in a standard normal dataset.
   Ans: The area of (-∞, -2.68] = 0.00368 = area of [2.68, ∞), so the proportion of mild outliers is 2 * 0.00368 = 0.00736 = 0.7%.
10. The heights of a population of 100,000 men are normally distributed with a mean of 68 inches with SD = 3. About how many men have heights
    
    1. below 65 inches?
    2. above 74 inches?
    3. between 62 and 74 inches?
11. IQ scores are normally distributed with mean 100 and standard deviation 15. Find the IQ scores that correspond to these percentiles:
         a. 50   Ans: 100
         b. 85   Ans: area of (-∞, z] = 0.85, so z = 1.04. This means that the IQ score is 1.04 SDs above average: the IQ score is 100 + 1.04 * 15 = 115.6
         c. 95   Ans: area of (-∞, z] = 0.95, so z = 1.64. This means that the IQ score is 1.64 SDs above average: the IQ score is 100 + 1.64 * 15 = 126.6
         d. 99.9   Ans: area of (-∞, z] = 0.999, so z = 3.08. This means that the IQ score is 3.08 SDs above average: the IQ score is 100 + 3.08 * 15 = 146.2