6.3: Finding Probabilities for the Normal Distribution

1. Male foot lengths have a normal distribution, with $\mu =11,\sigma =1.5$ inches. What is the probability of a foot length of more than 13 inches?

   
   First, we standardize: [latex]\mathcal{z}=\frac{\mathcal{x}-\mu}{\sigma}=\frac{13-11}{1.5}=+1.33[/latex]; the probability that we seek, P(X > 13), is the same as the probability

   P(Z > +1.33) that a normal variable takes a value greater than 1.33 standard deviations above its mean.

   

   This can be solved with the normal table, after applying the property of symmetry:

   P(Z > +1.33) = P(Z < -1.33) = .0918. A male foot length of more than 13 inches is on the long side, but not too unusual: its probability is about 9%.

   Comment:

   We can streamline the solution in terms of probability notation. Since the standardized value for 13 is (13 – 11) / 1.5 = +1.33, we can write $P(X>13)=P(Z>1.33)=P(Z<-1.33)=0.0918$.

   The first equality above holds because we subtracted the mean from a normal variable X and divided by its standard deviation, transforming it to a standardized normal variable that we call “Z.”

   The second equality above holds by the symmetry of the standard normal curve around zero.

   The last equality above was obtained from the normal table.

2. What is the probability of a male foot length between 10 and 12 inches?

   

   The standardized values of 10 and 12 are, respectively, [latex]\frac{10-11}{1.5}=-0.67[/latex] and [latex]\frac{12-11}{1.5}=0.67[/latex]

   

   P(-.67 < Z < +.67) = P(Z < +.67) – P(Z < -.67) = .7486 – .2514 = .4972.

   Or, if you prefer the streamlined notation,

   $P(10<X<12)=P(-0.67<Z<+0.67)=P(Z<+0.67)-P(Z<-0.67)=0.7486-0.2514=0.4972$.

Length (in days) of a randomly chosen human pregnancy is a normal random variable with $\mu =266,\sigma =16$

1. Find Q1, the median, and Q3. Q1 = 266 – .67(16) = 255; median = mean = 266;

   Q3 = 266 + .67(16) = 277. Thus, the probability is 1/4 that a pregnancy will last less than 255 days; 1/2 that it will last less than 266 days; 3/4 that it will last less than 277 days.

   

2. What is the probability that a randomly chosen pregnancy will last less than 246 days? Since (246 – 266) / 16 = -1.25, we write

   $P(X<246)=P(Z<-1.25)=0.1056$

3. What is the probability that a randomly chosen pregnancy will last longer than 240 days? Since (240 – 266) / 16 = -1.63, we write

   $P(X>240)=P(Z>-1.63)=P(Z<+1.63)=0.9484$

   Since the mean is 266 and the standard deviation is 16, most pregnancies last longer than 240 days.

4. What is the probability that a randomly chosen pregnancy will last longer than 500 days?

   Method 1: Common sense tells us that this would be impossible.

   Method 2: The standardized value of 500 is (500 – 266) / 16 = +14.625. $P(X>500)=P(Z>14.625)=0$.

5. Suppose a pregnant woman’s husband has scheduled his business trips so that he will be in town between the 235th and 295th days. What is the probability that the birth will take place during that time? The standardized values are (235 – 266) / 16) = -1.94 and (295 – 266) / 16 = +1.81.

   $P(235<X<295)=P(-1.94<Z<+1.81)=P(Z<+1.81)-P(Z<-1.94)=0.9649-0.0262=0.9387$

   There is close to a 94% chance that the husband will be in town for the birth.