Explore problems and real-world applications of normal distributions, complete with detailed solutions. To verify your answers, you can use our online normal probability calculator or inverse normal probability calculator.
Note: In this context, "area" refers to the area under the standard normal curve, and \( z \) represents the z-score, defined as:
\[ z = \dfrac{x - \mu}{\sigma} \]\( X \) is a normally distributed variable with mean \( \mu = 30 \) and standard deviation \( \sigma = 4 \). Find the probabilities
a) \( P(X \lt 40)\)
b) \( P(X \gt 21)\)
c) \( P(30 \lt X \lt 35) \)
a) For \( x = 40\) , the z-value \[ z = \dfrac {x - \mu}{\sigma } = \dfrac{40 - 30}{4} = 2.5 \] Hence \[ P(x \lt 40) = P(z \lt 2.5) = \text{area to the left of} \; 2.5 = 0.9938\]
b) For \( x = 21\), \[ z = \dfrac{21 - 30}{4} = -2.25 \] Hence \[ P(x \gt 21) = P(z \gt -2.25) \] \[ = \text{total area} - \text{area to the left of} -2.25 \] \[ = 1 - 0.0122 = 0.9878 \]
c) For \( x = 30 \) , \[ z = \dfrac{30 - 30}{4} = 0 \] and for \( x = 35 \), \[ z = \dfrac {35-30}{4} = 1.25\] Hence \[ P(30 \lt x \lt 35) = P(0 \lt z \lt 1.25) \] \[ = \text{area to the left of} z = 1.25] - \text{area to the left of} 0 \] \[ = 0.8944 - 0.5 = 0.3944 \]
A radar unit is used to measure the speeds of cars on a motorway. The speeds are normally distributed with a mean of \( 90 \) km/hr and a standard deviation of \( 10 \) km/hr. What is the probability that a car picked at random is traveling at more than \( 100 \) km/hr?
Let \( x \) be the random variable that represents the speed of cars. \( x \) has a mean \( \mu = 90 \) and a standard deviation \(\sigma = 10\).
We have to find the probability that \( x \) is higher than \( 100 \) or \( P(x \gt 100) \)
For \( x = 100\) , \[ z = \dfrac{100 - 90}{10} = 1 \] Hence \[ P(x \gt 90) = P(z \gt 1) \] \[ = \text{total area} - \text{area to the left of z = } 1 \] \[ = 1 - 0.8413 = 0.1587 \] The probability that a car selected at random has a speed greater than 100 km/hr is equal to 0.1587
For certain types 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 between charges will be between \( 50 \) and \( 70 \) hours.
Let \( x \) be the random variable that represents the length of time that 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 \( P( 50 \lt x \lt 70) \) For x = 50 , \[ z = \dfrac{50 - 50}{15} = 0 \] For x = 70 , \[ z = \dfrac{70-50}{15} = 1.3333 \quad \text{(rounded to 4 decimal places)} \] \[ P( 50 \lt x \lt 70) = P( 0 \lt z \lt 1.3333) \] \[ = \text{area to the left of z =} 1.3333] - \text{area to the left of z} = 0] \] \[ = 0.9088 - 0.5 = 0.4088 \] The probability that John's computer has a length of time between \( 50 \) and \( 70 \) hours is equal to \( 0.4088\).
Entry to a certain University is determined by a national test. The scores on this test are normally distributed with a mean of \( 500 \) and a standard deviation of \( 100 \). Tom wants to be admitted to this university and he knows that he must score better than at least \( 70\% \) of the students who took the test. Tom takes the test and scores \( 585 \). Will he be admitted to this university ?
Let \( x \) be the random variable that represents the test 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 \( x = 585 \) , \[ z = \dfrac{585 - 500}{100} = 0.85 \] The proportion \( P \) of students who scored below \( 585 \) is given by \[ P = \text{area to the left of z =} = 0.85 = 80.23\% \] Tom scored better than \( 80.23\% \) of the students who took the test and he will be admitted to this University.
The length of similar components produced by a company is 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 ?
b) What is the probability that the length of this component is between \( 4.96 \) and \( 5.04 \) cm?
a)
We need to calculate \[ P(4.98 \lt x \lt 5.02) \] For \( x = 4.98 \) \[ z = \dfrac{4.98 - 5}{0.02} = -1 \] For \( x = 5.02 \) \[ z = \dfrac{5.02 - 5}{0.02} = 1 \] Hence \[ P(4.98 \lt x \lt 5.02) = P(-1 \lt z \lt 1 \] \[ = 0.6826 \] b) We need to calculate \[ P(4.96 \lt x \lt 5.04) \] Calculte the z scores for \( x = 4.96 \) and \( x = 5.04 \) For \( x = 4.96 \) \[ z = \dfrac{4.96 - 5}{0.02} = - 2 \] For \( x = 5.04 \) \[ z = \dfrac{5.04 - 5}{0.02} = 2 \] Hence \[ P(4.96 \lt x \lt 5.04) = P(-2 \lt z \lt 2) \] \[ = 0.9544 \]
The length of life of an instrument produced by a machine has a normal distribution with a mean of \( 12 \) months and a standard deviation of \( 2 \) months. Find the probability that an instrument produced by this machine will last
a) less than 7 months.
b) between 7 and 12 months.
a) Find the z scores for \( x = 7 \), \[ z = \dfrac{7 - 12}{2} = 2.5 \] Hence \[ P(x \lt 7) = P(z \lt -2.5) = 0.0062\] b) Find the z scores for x = 12, \[ z = \dfrac{12 - 12}{2} = 0 \] Hence \[ P(7 \lt x \lt 12) = P(-2.5 \lt z \lt 0) = 0.4938 \]
The time taken to assemble a car in a certain plant is a random variable having a normal distribution with a mean \( 20 \) 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 ?
b) between 20 and 22 hours ?
a) Find the z scores for \( x = 19.5\), \[ z = \dfrac{19.5 - 20}{2} = -0.25 \] \[ P(x < 19.5) = P(z < -0.25) \] \[ = 0.4013 \] b) Find the z scores for \( x = 20\), \[ z = \dfrac{20 - 20}{2} = 0\] for \( x = 22\), \[ z = \dfrac{22 - 20}{2} = 1\] Hence \[ P(20 \lt x \lt 22) = P(0 \lt z \lt 1) = 0.3413 \]
A large group of students took a test in Physics and the final grades have a mean of 70 and a standard deviation of 10. If we can approximate the distribution of these grades by a normal distribution, what percent of the students
a) scored higher than \( 80 \) ?
b) Should pass the test, grades greater \( 60 \) ?
c) Should fail the test, grades less \( 60 \) ?
a) Find the z scores for \( x = 80\), \[ z = \dfrac{80 - 70}{10} = 1 \] Area to the right (higher than 80 = Area to the right (higher than) \( z = 1 \) and is equal to 0.1586
Hence \( 15.87 \% \) scored more that \( 80 \).
b) Find the z scores for \( x = 60\), \[ z = \dfrac{60 - 70}{10} = - 1 \] Area to the right (higher than) 60 = The area to the right of \( z = -1 \) is equal to \( 0.8413 \) = 84.13% should pass the test. Hence the percentage of students who should pass the test is \[ 84.13\% \] c) The percentage of students who should pass the test is is equal to percentage of those who should fail subtracted from \( 100 \% \) \[ 100\% - 15.87\% = 84.13 \$\]
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 \( \$40,000 \) ?
b) What percent of people earn between \( \$ 45,000 \) and \( \$ 65,000 \) ?
c) What percent of people earn more than \( \$ $70,000 \) ?
a)
Find the z scores for x = 40,000, z = -0.5 \[ z = \dfrac{40,000 - 50,000}{20} = -0.5 \] The percent of people earning less than 40,000 is the area to the left of of z = -0.5 is equal to \[ 0.3085 = 30.85\% \]
b) For \( x = 45,000 \) \[ z = \dfrac{45,000 - 50,000}{20} = -0.25 \] For \( x = 65,000 \) \[ z = \dfrac{65,000 - 50,000}{20} = 0.75 \] The percentage of people earning between \( \$ 45,000 \) and \( \$ 65,000 \) is the area between \( z = -0.25 \) and \( z = 0.75 \) is equal to \[ 0.3720 = 37.20 = 0.372\% \]
c) For x = 70000, \[ z = \dfrac{70,000 - 50,000}{20} = 1 \] The percentage of people earning more than \( \$ 70,000 \) is The area to the right of z = 1 and is equal to \[ 0.1586 = 15.86\% \]