15. The Normal Distribution

Question 1

From Section 9.5 we know the height of the uniform density is 1/40. The probability to the right of a particular value x is

P(Xx) = base × height = (190 - x) × (1/40).

From this we can build the following probability table:

Digits of x
x0123456789
1501.000.975.950.925.900.875.850.825.800.775
160.750.725.700.675.650.625.600.575.550.525
170.500.475.450.425.400.375.350.325.300.275
180.250.225.200.175.150.125.100.075.050.025

 

Question 2

From Section 9.5 we know the height of the triangular density is 1/20. If x is 170 or more then the height of the density curve at x is (1/20)×((190 - x)/20). You can check that this gives height 0 for x=190 and height 1/20 for x=170. The area to the right of x is then

P(Xx) = (1/2) × base × height = (1/2) × (190 - x) × (1/20)×((190 - x)/20) = (1/800)×(190 - x)2.

Similarly, for x ≤ 170 the area to the right is

P(Xx) = 1 - (1/800)×(x - 150)2.

From these we can build the following probability table:

Digits of x
x0123456789
1501.000.999.995.989.980.969.955.939.920.899
160.875.849.820.789.755.719.680.639.595.549
170.500.451.405.361.320.281.245.211.180.151
180.125.101.080.061.045.031.020.011.005.001

 

Question 3

Unlike the previous two questions, for Normal probabilities it is not possible to write down a formula for the area to the right of a particular x. Instead we will standardize our x values, using μ = 167 and σ = 6.6, and then use the table of standard Normal probabilities (Table 15.1).

For example, to find P(X ≥ 170) we calculate z = (170 - 167)/6.6 = 0.45. Table 15.1 then gives P(Z ≥ 0.45) = .326.

Digits of x
x0123456789
1500.9950.9920.9880.9830.9760.9660.9530.9360.9130.887
1600.8550.8190.7760.7290.6740.6180.5600.5000.4400.382
1700.3260.2710.2240.1810.1450.1130.0870.0640.0470.034
1800.0240.0170.0120.0080.0050.0030.0020.0010.0010.000

 

Question 4

Four boxplots are shown below. There is certainly a lot of variability in the kind of picture you get from a boxplot when the sample size is so small.