Applications of the Normal Distribution

• Some of the many problems for which the normal distribution is applicable are treated in the following examples.
• Example 6.7: A certain type of storage battery lasts, on average, 3.0 years, with a standard deviation of 0.5 year.
• Assuming that the battery lives are normally distributed, find the probability that a given battery will last less than 2.3 years.
• Solution:
  $$z=\frac{2.3-3}{0.5}=-1.4 \Rightarrow P(X<2.3)=P(Z<-1.4)=0.0808$$

  Figure 6.14: Area for Example 6.7.

  

• Example 6.8: An electrical firm manufactures light bulbs that have a life, before burn-out, that is normally distributed with mean equal to 800 hours and a standard deviation of 40 hours.
• Find the probability that a bulb burns between 778 and 834 hours.
• Solution:
  $$z_1=\frac{778-800}{40}=-0.55~and~z_2=\frac{834-800}{40}=0.85$$
  $$P(778< X < 834)=P(-0.55 < Z <0.85)$$
  $$=P(Z<0.85)-P(Z<-0.55)=0.8023-0.2912=0.5111$$

  Figure 6.15: Area for Example 6.8.

  

• Example 6.9: The buyer sets specifications on the diameter to be $3.0 \pm 0.01$ cm.
• It is known that in the process the diameter of a ball bearing has a normal distribution with mean $\mu=3.0$ and standard deviation $\sigma=0.005$.
• On the average, how many manufactured ball bearings will be scrapped?.
• Solution:
  $$z_1=\frac{2.99-3.0}{0.005}=-2.0$$
  $$z_2=\frac{3.01-3.0}{0.005}=2.0$$
  $$\Rightarrow P(2.99 < X < 3.01)$$
  $$=P(-2.0 < Z <2.0)$$
  $$=1-2*P(Z<-2.0)$$
  $$=1-2*0.0228=0.9544$$

  Figure 6.16: Area for Example 6.9.

  

• Example 6.10: Gauges are used to reject all components where a certain dimension is not within the specification $1.50 \pm d$.
• It is known that this measurement is normally distributed with mean $\mu=1.50$ and standard deviation $\sigma=0.2$.
• Determine the value $d$ such that the specifications cover 95% of the measurements.
• Solution:

  From Table A.3 we know that

  $$P(-1.96 < Z < 1.96)=0.95$$
  $$1.96=\frac{(1.50+d)-1.50}{0.2}$$
  $$\Rightarrow d=0.2*1.96=0.392$$

  Figure 6.17: Specifications for Example 6.10.

  

• Example 6.11: A certain machine makes electrical resistors having a mean resistance of 40 ohms and a standard deviation of 2 ohms.
• Assuming that the resistance follows a normal distribution and can be measured to any degree of accuracy, what percentage of resistors will have a resistance exceeding 43 ohms?
• Solution:

  From Table A.3 we know that

  $$z=\frac{43-40}{2}=1.5$$
  $$P(X>43)=P(Z>1.5)$$
  $$=1-P(Z<1.5)=1-0.9332$$
  $$=0.0668$$

  Figure 6.18: Area for Example 6.11.

  

• Example 6.13: The average grade for an exam is 74, and the standard deviation is 7.
• If 12% of the class are given $A$'s, and the grades are curved to follow a normal distribution,
• What is the lowest possible $A$ and the highest possible $B$?
• Solution:
  $$1-0.12=0.88 =P(Z<1.175)$$
  $$1.175=\frac{x-74}{7} \Rightarrow$$
  $$x=7*1.175+74=82.225$$
  The lowest $A$ is 83 and the highest $B$ is 82.

  Figure 6.19: Area for Example 6.13.