Sample Probability Problems using Normal Distribution Calculator

Note: this is a draft document- work in progress. What is here is correct but needs polish and citations.
Background review of probability and the Normal curve: The normal distribution is a continuous probability distribution. This has several implications for probability.

  • The total area under the normal curve is equal to 1.
  • The probability that a normal random variable X equals any particular value is 0.
  • The probability that X is greater than a equals the area under the normal curve bounded by a and plus infinity (as indicated by the non-shaded area in the figure below).
  • The probability that X is less than a equals the area under the normal curve bounded by a and minus infinity (as indicated by the shaded area in the figure below).

    Normal Probability curve

    Normal Probability curve

 

 

 

 

 

Sample problems using Normal Distribution Calculator here

1. X is a normally normally distributed variable with mean μ = 20 and standard deviation σ = 4. Find:
a) P(x < 30). Remember that a Z-distribution has a mean of 0 and a S.D. of 1. For x = 30, the z-value z = (30 – 20) / 4 = 2.5. Hence P(x < 30) = P(z < 2.5) = [area to the left of 2.5] = 0.9938.
b) P(x > 21). For x = 11, z = (11 – 20) / 4 = -2.25. Hence P(x > 21) = P(z > -2.25) = [total area] – [area to the left of -2.25] = 1 – 0.0122 = 0.9878.
c) P(20 < x < 25). For x = 20 , z = (20 – 20) / 4 = 0 and for x = 25, z = (25 – 20) / 4 = 1.25. Hence P(20 < x < 25) = P(0 < z < 1.25) = [area to the left of z = 1.25] – [area to the left of 0] = 0.8944 – 0.5 = 0.3944.
2. A radar unit is used to measure speeds of cars on a motorway. The speeds are normally distributed with a mean of 60 mi/hr and a standard deviation of 10 mi/hr. What is the probability that a car picked at random is travelling at more than 70 km/hr?
Let x be the random variable that represents the speed of cars. x has μ = 60 and σ = 10. We have to find the probability that x is higher than 70 or P(x > 70). For x = 100 , z = (70 – 60) / 10 = 1. P(x > 70) = P(z >, 1) = [total area] – [area to the left of z = 1] = 1 – 0.8413 = 0.1587. The probability that a car selected at a random has a speed greater than 100 km/hr is equal to 0.1587.
3. For a certain type of computers, the length of time between charges of the battery is normally distributed with a mean of 50 hours and a standard deviation of 15 hours. John owns one of these computers and wants to know the probability that the length of time will be between 50 and 70 hours.
Let x be the random variable that represents the length of time. It has a mean of 50 and a standard deviation of 15. We have to find the probability that x is between 50 and 70 or P( 50< x < 70). For x = 50 , z = (50 – 50) / 15 = 0 For x = 70 , z = (70 – 50) / 15 = 1.33 (rounded to 2 decimal places) P( 50< x < 70) = P( 0< z < 1.33) = [area to the left of z = 1.33] – [area to the left of z = 0] = 0.9082 – 0.5 = 0.4082 The probability that John’s computer has a length of time between 50 and 70 hours is equal to 0.4082.
4. Entry to a certain college is determined by a national test. The scores on this test are normally distributed with a mean of 500 and a standard deviation of 100. Joe wants to be admitted to this college and he knows that he must score better than at least 70% of the students who took the test. Joe takes the test and scores 585. Will he be admitted to this college?
Let x be the random variable that represents the scores. x is normally distributed with a mean of 500 and a standard deviation of 100. The total area under the normal curve represents the total number of students who took the test. If we multiply the values of the areas under the curve by 100, we obtain percentages. For x = 585 , z = (585 – 500) / 100 = 0.85 .  The proportion P of students who scored below 585 is given by P = [area to the left of z = 0.85] = 0.8023 = 80.23%. Joe scored better than 80.23% of the students who took the test and he will be admitted to this college.
5. The length of similar components produced by Acme Pipe are approximated by a normal distribution model with a mean of 5 cm and a standard deviation of 0.02 cm. If a component is chosen at random :
a) what is the probability that the length of this component is between 4.98 and 5.02 cm? 4.98 and 5.02 cm represent 1 SD above and below the mean. Therefore P(4.98 < x < 5.02) = P(-1 < z < 1) = 0.6826.
b) What is the probability that the length of this component is between 4.96 and 5.04 cm? P(4.96 < x < 5.04) = P(-2 < z < 2) = 0.9544
6. The length of life of an instrument produced by a machine has a normal distribution with a mean of 12 months and standard deviation of 2 months. Find the probability that an instrument produced by this machine will last
a) less than 7 months. P(x < 7) = P(z < -2.5) = 0.0062
b) between 7 and 12 months. P(7 < x < 12) = P(-2.5 < z < 0) = 0.4938
7. The time taken to assemble a car in a certain plant is a random variable having a normal distribution of 20 hours and a standard deviation of 2 hours. What is the probability that a car can be assembled at this plant in a period of time
a) less than 19.5 hours? P(x < 19.5) = P(z < -0.25) = 0.4013
b) between 20 and 22 hours? P(20 < x < 22) = P(0 < z < 1) = 0.3413
8. A large group of students took a test in Chemistry and the final grades have a mean of 70 and a standard deviation of 10. Assuming a normal distribution, what percent of the students
a) scored higher than 80? For x = 80, z = 1. Area to the right (higher than) z = 1 is equal to 0.1586 = 15.87% scored more that 80.
b) should pass the test (grades60)? For x = 60, z = -1 . Area to the right of z = -1 is equal to 0.8413 = 84.13% should pass the test.
c) should fail the test (grades<60)? 100% – 84.13% = 15.87% should fail the test.
9. The annual salaries of employees in a large company are approximately normally distributed with a mean of $50,000 and a standard deviation of $20,000.
a) What percent of people earn less than $40,000? For x = 40000, z = -0.5. Area to the left (less than) of z = -0.5 is equal to 0.3085 = 30.85% earn less than $40,000.
b) What percent of people earn between $45,000 and $65,000? For x = 45000 , z = -0.25 and for x = 65000, z = 0.75. Area between z = -0.25 and z = 0.75 is equal to 0.3720 = 37.20% earn between $45,000 and $65,000.
c) What percent of people earn more than $70,000? For x = 70000, z = 1. Area to the right (higher) of z = 1 is equal to 0.1586 = 15.86% earn more than $70,000.

10.  An average light bulb manufactured by the Acme Corporation lasts 300 days with a standard deviation of 50 days. Assuming that bulb life is normally distributed, what is the probability that an Acme light bulb will last at most 365 days? Solution: Given a mean score of 300 days and a standard deviation of 50 days, we want to find the cumulative probability that bulb life is less than or equal to 365 days. Thus, we know the following:

  • The value of the normal random variable is 365 days.
  • The mean is equal to 300 days.
  • The standard deviation is equal to 50 days.

We enter these values into the Normal Distribution Calculator and compute the cumulative probability. The answer is: P( X < 365) = 0.90. Hence, there is a 90% chance that a light bulb will burn out within 365 days.   11. Suppose scores on an IQ test are normally distributed. If the test has a mean of 100 and a standard deviation of 10, what is the probability that a person who takes the test will score between 90 and 110? Solution: Here, we want to know the probability that the test score falls between 90 and 110. The “trick” to solving this problem is to realize the following:

P( 90 < X < 110 ) = P( X < 110 ) – P( X < 90 )

We use the Normal Distribution Calculator to compute both probabilities on the right side of the above equation.

  1. To compute P( X < 110 ), we enter the following inputs into the calculator: The value of the normal random variable is 110, the mean is 100, and the standard deviation is 10. We find that P( X < 110 ) is 0.84.
  2. To compute P( X < 90 ), we enter the following inputs into the calculator: The value of the normal random variable is 90, the mean is 100, and the standard deviation is 10. We find that P( X < 90 ) is 0.16.

We use these findings to compute our final answer as follows:

P( 90 < X < 110 ) = P( X < 110 ) – P( X < 90 ) P( 90 < X < 110 ) = 0.84 – 0.16 P( 90 < X < 110 ) = 0.68

Thus, about 68% of the test scores will fall between 90 and 110.