Sure, here are **20 practice problems related to the normal distribution of engineering mathematics** along with their solutions:

## Please note: Below each sum, a hint is given. Please solve the sum without watching the hind. In case you fail to solve, then view the hint. If still you are unable to solve it, then view the detailed solution given at the end of this set of problems.

1. **Problem**: The heights of students in a class follow a normal distribution with a mean of 65 inches and a standard deviation of 3 inches. What percentage of students have heights between 62 inches and 68 inches?

**Solution**: To find this, we calculate the z-scores for 62 and 68 using the formula z = (x – μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation. Then, we use a standard normal distribution table or calculator to find the probabilities and subtract them.

2. **Problem**: The diameter of bolts produced in a factory follows a normal distribution with a mean of 10 mm and a standard deviation of 0.2 mm. What is the probability that a randomly selected bolt has a diameter greater than 10.5 mm?

**Solution**: Similar to problem 1, we find the z-score for 10.5 mm, then use the standard normal distribution table or calculator to find the probability of a value being greater than that z-score.

3. **Problem**: The lifespans of batteries produced by a company follow a normal distribution with a mean of 4 years and a standard deviation of 0.5 years. What is the probability that a randomly selected battery will last between 3.5 and 4.5 years?

**Solution**: We find the z-scores for 3.5 and 4.5 years, then use the standard normal distribution table or calculator to find the probabilities and subtract them.

4. **Problem**: The braking distances of cars on a certain road follow a normal distribution with a mean of 50 meters and a standard deviation of 5 meters. What percentage of cars will stop within 45 and 55 meters?

**Solution**: Similar to problem 3, we find the z-scores for 45 and 55 meters, then use the standard normal distribution table or calculator to find the probabilities and subtract them.

5. **Problem**: The temperatures recorded in a city follow a normal distribution with a mean of 25°C and a standard deviation of 2°C. What is the probability that the temperature will exceed 28°C?

**Solution**: We find the z-score for 28°C, then use the standard normal distribution table or calculator to find the probability of a value being greater than that z-score.

6. **Problem**: The weights of chocolate bars produced in a factory follow a normal distribution with a mean of 100 grams and a standard deviation of 10 grams. What is the probability that a randomly selected chocolate bar will weigh less than 90 grams?

**Solution**: We find the z-score for 90 grams, then use the standard normal distribution table or calculator to find the probability of a value being less than that z-score.

7. **Problem**: The scores on a standardized test follow a normal distribution with a mean of 500 and a standard deviation of 100. What percentage of students scored above 600?

**Solution**: We find the z-score for 600, then use the standard normal distribution table or calculator to find the probability of a value being greater than that z-score.

8. **Problem**: The fill volumes of soda bottles follow a normal distribution with a mean of 500 ml and a standard deviation of 20 ml. What is the probability that a randomly selected bottle contains between 480 ml and 520 ml?

**Solution**: We find the z-scores for 480 and 520 ml, then use the standard normal distribution table or calculator to find the probabilities and subtract them.

9. **Problem**: The reaction times of drivers follow a normal distribution with a mean of 0.8 seconds and a standard deviation of 0.1 seconds. What percentage of drivers have reaction times faster than 0.9 seconds?

**Solution**: We find the z-score for 0.9 seconds, then use the standard normal distribution table or calculator to find the probability of a value being less than that z-score.

10. **Problem**: The salaries of employees in a company follow a normal distribution with a mean of $50,000 and a standard deviation of $5,000. What is the probability that a randomly selected employee earns less than $45,000?

**Solution**: We find the z-score for $45,000, then use the standard normal distribution table or calculator to find the probability of a value being less than that z-score.

## Now solve 7 more problems:

11. **Problem**: The lengths of screws produced in a factory follow a normal distribution with a mean of 25 mm and a standard deviation of 2 mm. What is the probability that a randomly selected screw will be longer than 28 mm?

**Solution**: We find the z-score for 28 mm, then use the standard normal distribution table or calculator to find the probability of a value being greater than that z-score.

12. **Problem**: The IQ scores of a population follow a normal distribution with a mean of 100 and a standard deviation of 15. What percentage of the population has an IQ between 85 and 115?

**Solution**: We find the z-scores for 85 and 115, then use the standard normal distribution table or calculator to find the probabilities and subtract them.

13. **Problem**: The speeds of cars on a highway follow a normal distribution with a mean of 70 mph and a standard deviation of 5 mph. What is the probability that a randomly selected car is traveling at more than 75 mph?

**Solution**: We find the z-score for 75 mph, then use the standard normal distribution table or calculator to find the probability of a value being greater than that z-score.

14. **Problem**: The heights of sunflowers in a field follow a normal distribution with a mean of 150 cm and a standard deviation of 10 cm. What percentage of sunflowers have heights between 140 cm and 160 cm?

**Solution**: We find the z-scores for 140 and 160 cm, then use the standard normal distribution table or calculator to find the probabilities and subtract them.

15. **Problem**: The reaction times of athletes follow a normal distribution with a mean of 0.2 seconds and a standard deviation of 0.05 seconds. What is the probability that a randomly selected athlete has a reaction time slower than 0.25 seconds?

**Solution**: We find the z-score for 0.25 seconds, then use the standard normal distribution table or calculator to find the probability of a value being less than that z-score.

16. **Problem**: The diameters of bearings produced in a factory follow a normal distribution with a mean of 50 mm and a standard deviation of 2 mm. What is the probability that a randomly selected bearing has a diameter between 48 mm and 52 mm?

**Solution**: We find the z-scores for 48 and 52 mm, then use the standard normal distribution table or calculator to find the probabilities and subtract them.

17. **Problem**: The blood pressure readings of patients follow a normal distribution with a mean of 120 mmHg and a standard deviation of 10 mmHg. What percentage of patients have blood pressure readings above 130 mmHg?

**Solution**: We find the z-score for 130 mmHg, then use the standard normal distribution table or calculator to find the probability of a value being greater

Read free essay that are suitable for students.

## See the Detail Solutions:

**Problem 1:** The heights of students in a class follow a normal distribution with a mean of 65 inches and a standard deviation of 3 inches. What percentage of students have heights between 62 inches and 68 inches?

Solution:

Given:

– Mean (μ) = 65 inches

– Standard deviation (σ) = 3 inches

– Lower height limit (x₁) = 62 inches

– Upper height limit (x₂) = 68 inches

First, let’s find the z-scores for both limits using the formula:

\[ z = \frac{x – \mu}{\sigma} \]

For \( x₁ = 62 \) inches:

\[ z₁ = \frac{62 – 65}{3} = \frac{-3}{3} = -1 \]

For \( x₂ = 68 \) inches:

\[ z₂ = \frac{68 – 65}{3} = \frac{3}{3} = 1 \]

Now, we find the probabilities corresponding to these z-scores using a standard normal distribution table or calculator.

From the standard normal distribution table, we find:

– \( P(Z < -1) \approx 0.1587 \)

– \( P(Z < 1) \approx 0.8413 \)

To find the probability between the two limits, we subtract the cumulative probability corresponding to \( z₁ \) from the cumulative probability corresponding to \( z₂ \):

\[ P(-1 < Z < 1) = P(Z < 1) – P(Z < -1) = 0.8413 – 0.1587 = 0.6826 \]

Therefore, approximately 68.26% of students have heights between 62 inches and 68 inches.

**Problem 2:** The diameter of bolts produced in a factory follows a normal distribution with a mean of 10 mm and a standard deviation of 0.2 mm. What is the probability that a randomly selected bolt has a diameter greater than 10.5 mm?

Solution:

Given:

– Mean (μ) = 10 mm

– Standard deviation (σ) = 0.2 mm

– Desired diameter (x) = 10.5 mm

First, let’s find the z-score using the formula:

\[ z = \frac{x – \mu}{\sigma} \]

Substituting the given values:

\[ z = \frac{10.5 – 10}{0.2} = \frac{0.5}{0.2} = 2.5 \]

Now, we find the probability of a z-score being greater than 2.5 using a standard normal distribution table or calculator.

From the standard normal distribution table, we find:

– \( P(Z > 2.5) \approx 1 – 0.9938 \approx 0.0062 \)

Therefore, the probability that a randomly selected bolt has a diameter greater than 10.5 mm is approximately 0.0062 or 0.62%.

Let me know if you’d like me to continue solving the remaining problems!

Certainly! Let’s continue solving the remaining problems:

**Problem 3:** The lifespans of batteries produced by a company follow a normal distribution with a mean of 4 years and a standard deviation of 0.5 years. What is the probability that a randomly selected battery will last between 3.5 and 4.5 years?

Solution:

Given:

– Mean (μ) = 4 years

– Standard deviation (σ) = 0.5 years

– Lower lifespan limit (x₁) = 3.5 years

– Upper lifespan limit (x₂) = 4.5 years

First, let’s find the z-scores for both limits using the formula:

\[ z = \frac{x – \mu}{\sigma} \]

For \( x₁ = 3.5 \) years:

\[ z₁ = \frac{3.5 – 4}{0.5} = -1 \]

For \( x₂ = 4.5 \) years:

\[ z₂ = \frac{4.5 – 4}{0.5} = 1 \]

Now, we find the probabilities corresponding to these z-scores using a standard normal distribution table or calculator.

From the standard normal distribution table, we find:

– \( P(Z < -1) \approx 0.1587 \)

– \( P(Z < 1) \approx 0.8413 \)

To find the probability between the two limits, we subtract the cumulative probability corresponding to \( z₁ \) from the cumulative probability corresponding to \( z₂ \):

\[ P(-1 < Z < 1) = P(Z < 1) – P(Z < -1) = 0.8413 – 0.1587 = 0.6826 \]

Therefore, approximately 68.26% of batteries will last between 3.5 and 4.5 years.

**Problem 4:** The braking distances of cars on a certain road follow a normal distribution with a mean of 50 meters and a standard deviation of 5 meters. What percentage of cars will stop within 45 and 55 meters?

Solution:

Given:

– Mean (μ) = 50 meters

– Standard deviation (σ) = 5 meters

– Lower distance limit (x₁) = 45 meters

– Upper distance limit (x₂) = 55 meters

First, let’s find the z-scores for both limits using the formula:

\[ z = \frac{x – \mu}{\sigma} \]

For \( x₁ = 45 \) meters:

\[ z₁ = \frac{45 – 50}{5} = -1 \]

For \( x₂ = 55 \) meters:

\[ z₂ = \frac{55 – 50}{5} = 1 \]

Now, we find the probabilities corresponding to these z-scores using a standard normal distribution table or calculator.

From the standard normal distribution table, we find:

– \( P(Z < -1) \approx 0.1587 \)

– \( P(Z < 1) \approx 0.8413 \)

To find the probability between the two limits, we subtract the cumulative probability corresponding to \( z₁ \) from the cumulative probability corresponding to \( z₂ \):

\[ P(-1 < Z < 1) = P(Z < 1) – P(Z < -1) = 0.8413 – 0.1587 = 0.6826 \]

Therefore, approximately 68.26% of cars will stop within 45 and 55 meters.

Would you like me to continue solving the remaining problems?

Certainly! Let’s solve the 5th problem:

**Problem 5:** The temperatures recorded in a city follow a normal distribution with a mean of 25°C and a standard deviation of 2°C. What is the probability that the temperature will exceed 28°C?

Solution:

Given:

– Mean (μ) = 25°C

– Standard deviation (σ) = 2°C

– Desired temperature (x) = 28°C

First, let’s find the z-score using the formula:

\[ z = \frac{x – \mu}{\sigma} \]

Substituting the given values:

\[ z = \frac{28 – 25}{2} = \frac{3}{2} = 1.5 \]

Now, we find the probability of a z-score being greater than 1.5 using a standard normal distribution table or calculator.

From the standard normal distribution table, we find:

– \( P(Z > 1.5) \approx 1 – 0.9332 \approx 0.0668 \)

Therefore, the probability that the temperature will exceed 28°C is approximately 0.0668 or 6.68%.

This indicates that there is about a 6.68% chance that the temperature recorded in the city will be higher than 28°C.

Certainly! Let’s continue solving the problems:

Problem 7: The scores on a standardized test follow a normal distribution with a mean of 500 and a standard deviation of 100. What percentage of students scored above 600?

Solution:

Given:

– Mean (μ) = 500

– Standard deviation (σ) = 100

– Desired score (x) = 600

First, let’s find the z-score using the formula:

\[ z = \frac{x – \mu}{\sigma} \]

Substituting the given values:

\[ z = \frac{600 – 500}{100} = \frac{100}{100} = 1 \]

Now, we find the probability of a z-score being greater than 1 using a standard normal distribution table or calculator.

From the standard normal distribution table, we find:

– \( P(Z > 1) \approx 1 – 0.8413 \approx 0.1587 \)

Therefore, the probability that a student scored above 600 is approximately 0.1587 or 15.87%.

Problem 8: The fill volumes of soda bottles follow a normal distribution with a mean of 500 ml and a standard deviation of 20 ml. What is the probability that a randomly selected bottle contains between 480 ml and 520 ml?

Solution:

Given:

– Mean (μ) = 500 ml

– Standard deviation (σ) = 20 ml

– Lower volume limit (x₁) = 480 ml

– Upper volume limit (x₂) = 520 ml

We first find the z-scores for both limits:

\[ z₁ = \frac{480 – 500}{20} = -1 \]

\[ z₂ = \frac{520 – 500}{20} = 1 \]

Now, we find the probabilities corresponding to these z-scores:

– \( P(Z < -1) \approx 0.1587 \)

– \( P(Z < 1) \approx 0.8413 \)

To find the probability between the two limits:

\[ P(-1 < Z < 1) = P(Z < 1) – P(Z < -1) = 0.8413 – 0.1587 = 0.6826 \]

Therefore, approximately 68.26% of soda bottles contain between 480 ml and 520 ml.

Problem 9: The reaction times of drivers follow a normal distribution with a mean of 0.8 seconds and a standard deviation of 0.1 seconds. What percentage of drivers have reaction times faster than 0.9 seconds?

Solution:

Given:

– Mean (μ) = 0.8 seconds

– Standard deviation (σ) = 0.1 seconds

– Desired reaction time (x) = 0.9 seconds

We find the z-score for 0.9 seconds:

\[ z = \frac{0.9 – 0.8}{0.1} = 1 \]

Now, we find the probability of a z-score being greater than 1:

– \( P(Z > 1) \approx 1 – 0.8413 \approx 0.1587 \)

Therefore, approximately 15.87% of drivers have reaction times faster than 0.9 seconds.

**Problem 10:** The salaries of employees in a company follow a normal distribution with a mean of $50,000 and a standard deviation of $5,000. What is the probability that a randomly selected employee earns less than $45,000?

Solution:

Given:

– Mean (μ) = $50,000

– Standard deviation (σ) = $5,000

– Desired salary (x) = $45,000

We find the z-score for $45,000:

\[ z = \frac{45,000 – 50,000}{5,000} = -1 \]

Now, we find the probability of a z-score being less than -1:

– \( P(Z < -1) \approx 0.1587 \)

Therefore, the probability that a randomly selected employee earns less than $45,000 is approximately 0.1587 or 15.87%.

**Problem 11:** The lengths of screws produced in a factory follow a normal distribution with a mean of 25 mm and a standard deviation of 2 mm. What is the probability that a randomly selected screw will be longer than 28 mm?

Solution:

Given:

– Mean (μ) = 25 mm

– Standard deviation (σ) = 2 mm

– Desired length (x) = 28 mm

We find the z-score for 28 mm:

\[ z = \frac{28 – 25}{2} = 1.5 \]

Now, we find the probability of a z-score being greater than 1.5:

– \( P(Z > 1.5) \approx 1 – 0.9332 \approx 0.0668 \)

Therefore, the probability that a randomly selected screw will be longer than 28 mm is approximately 0.0668 or 6.68%.

Problem 12: The IQ scores of a population follow a normal distribution with a mean of 100 and a standard deviation of 15. What percentage of the population has an IQ between 85 and 115?

Solution:

Given:

– Mean (μ) = 100

– Standard deviation (σ) = 15

– Lower IQ limit (x₁) = 85

– Upper IQ limit (x₂) = 115

We first find the z-scores for both limits:

\[ z₁ = \frac{85 – 100}{15} = -1 \]

\[ z₂ = \frac{115 – 100}{15} = 1 \]

Now, we find the probabilities corresponding to these z-scores:

– \( P(Z < -1) \approx 0.1587 \)

– \( P(Z < 1) \approx 0.8413 \)

To find the probability between the two limits:

\[ P(-1 < Z < 1) = P(Z < 1) – P(Z < -1) = 0.8413 – 0.1587 = 0.6826 \]

Therefore, approximately 68.26% of the population has an IQ between 85 and 115.

**Problem 13:** The speeds of cars on a highway follow a normal distribution with a mean of 70 mph and a standard deviation of 5 mph. What is the probability that a randomly selected car is traveling at more than 75 mph?

Solution:

Given:

– Mean (μ) = 70 mph

– Standard deviation (σ) = 5 mph

– Desired speed (x) = 75 mph

We find the z-score for 75 mph:

\[ z = \frac{75 – 70}{5} = 1 \]

Now, we find the probability of a z-score being greater than 1:

– \( P(Z > 1) \approx 1 – 0.8413 \approx 0.1587 \)

Therefore, the probability that a randomly selected car is traveling at more than 75 mph is approximately 0.1587 or 15.87%.

Problem 14: The heights of sunflowers in a field follow a normal distribution with a mean of 150 cm and a standard deviation

of 10 cm. What percentage of sunflowers have heights between 140 cm and 160 cm?

Solution:

Given:

– Mean (μ) = 150 cm

– Standard deviation (σ) = 10 cm

– Lower height limit (x₁) = 140 cm

– Upper height limit (x₂) = 160 cm

We first find the z-scores for both limits:

\[ z₁ = \frac{140 – 150}{10} = -1 \]

\[ z₂ = \frac{160 – 150}{10} = 1 \]

Now, we find the probabilities corresponding to these z-scores:

– \( P(Z < -1) \approx 0.1587 \)

– \( P(Z < 1) \approx 0.8413 \)

To find the probability between the two limits:

\[ P(-1 < Z < 1) = P(Z < 1) – P(Z < -1) = 0.8413 – 0.1587 = 0.6826 \]

Therefore, approximately 68.26% of sunflowers have heights between 140 cm and 160 cm.

**Problem 15:** The reaction times of athletes follow a normal distribution with a mean of 0.2 seconds and a standard deviation of 0.05 seconds. What is the probability that a randomly selected athlete has a reaction time slower than 0.25 seconds?

Solution:

Given:

– Mean (μ) = 0.2 seconds

– Standard deviation (σ) = 0.05 seconds

– Desired reaction time (x) = 0.25 seconds

We find the z-score for 0.25 seconds:

\[ z = \frac{0.25 – 0.2}{0.05} = 1 \]

Now, we find the probability of a z-score being less than 1:

– \( P(Z < 1) \approx 0.8413 \)

Therefore, the probability that a randomly selected athlete has a reaction time slower than 0.25 seconds is approximately 0.8413 or 84.13%.

**Problem 16:** The diameters of bearings produced in a factory follow a normal distribution with a mean of 50 mm and a standard deviation of 2 mm. What is the probability that a randomly selected bearing has a diameter between 48 mm and 52 mm?

Solution:

Given:

– Mean (μ) = 50 mm

– Standard deviation (σ) = 2 mm

– Lower diameter limit (x₁) = 48 mm

– Upper diameter limit (x₂) = 52 mm

We first find the z-scores for both limits:

\[ z₁ = \frac{48 – 50}{2} = -1 \]

\[ z₂ = \frac{52 – 50}{2} = 1 \]

– \( P(Z < -1) \approx 0.1587 \)

– \( P(Z < 1) \approx 0.8413 \)

\[ P(-1 < Z < 1) = P(Z < 1) – P(Z < -1) = 0.8413 – 0.1587 = 0.6826 \]

Therefore, approximately 68.26% of bearings have diameters between 48 mm and 52 mm.