If bolt thread length is normally distributed, what is the probability that the thread length of a randomly selected bolt is

a. Within 1.5 SDs of its mean value?

b. Farther than 2.5 SDs from its mean value?

c. Between 1 and 2 SDs from its mean value?

We can solve each part of the problem using the properties of the standard normal distribution (a normal distribution with a mean of 0 and standard deviation of 1).

### a. Probability that the thread length is within 1.5 standard deviations of the mean:

The probability that a value is within 1.5 standard deviations of the mean corresponds to the area under the standard normal curve between \( -1.5 \) and \( 1.5 \). Using the standard normal table or a calculator, we find:

\[
P(-1.5 \leq Z \leq 1.5)
\]

First, find the cumulative probability for \( Z = 1.5 \) from a standard normal distribution:

\[
P(Z \leq 1.5) = 0.9332
\]

Since the normal distribution is symmetric around the mean:

\[
P(Z \geq -1.5) = P(Z \leq 1.5) = 0.9332
\]

Thus, the probability of being within 1.5 standard deviations of the mean is:

\[
P(-1.5 \leq Z \leq 1.5) = 0.9332 - (1 - 0.9332) = 0.9332 - 0.0668 = 0.8664
\]

So, the probability is **0.8664**, or **86.64%**.

### b. Probability that the thread length is farther than 2.5 standard deviations from the mean:

This is the probability that the value lies outside the range of \( -2.5 \) to \( 2.5 \). The formula would be:

\[
P(Z < -2.5 \text{ or } Z > 2.5) = 1 - P(-2.5 \leq Z \leq 2.5)
\]

Find the cumulative probability for \( Z = 2.5 \):

\[
P(Z \leq 2.5) = 0.9938
\]

Thus, the probability of being within 2.5 standard deviations of the mean is:

\[
P(-2.5 \leq Z \leq 2.5) = 0.9938
\]

Therefore, the probability of being farther than 2.5 standard deviations from the mean is:

\[
P(Z < -2.5 \text{ or } Z > 2.5) = 1 - 0.9938 = 0.0062
\]

So, the probability is **0.0062**, or **0.62%**.

### c. Probability that the thread length is between 1 and 2 standard deviations from the mean:

This is the probability that the value lies between \( 1 \) and \( 2 \) standard deviations. The formula is:

\[
P(1 \leq Z \leq 2)
\]

Find the cumulative probabilities for \( Z = 1 \) and \( Z = 2 \):

\[
P(Z \leq 1) = 0.8413
\]
\[
P(Z \leq 2) = 0.9772
\]

Thus, the probability of being between 1 and 2 standard deviations is:

\[
P(1 \leq Z \leq 2) = P(Z \leq 2) - P(Z \leq 1) = 0.9772 - 0.8413 = 0.1359
\]

So, the probability is **0.1359**, or **13.59%**.

### Summary:
- a. Probability within 1.5 SDs: **0.8664 (86.64%)**
- b. Probability farther than 2.5 SDs: **0.0062 (0.62%)**
- c. Probability between 1 and 2 SDs: **0.1359 (13.59%)**

Would you like more details on how I calculated these probabilities?

Here are 5 related questions:
1. What is the probability of being within 2 standard deviations of the mean?
2. How would the probabilities change if the standard deviation increased?
3. What is the probability that a bolt thread length is more than 3 SDs from the mean?
4. How do you calculate z-scores for normally distributed data?
5. How does the cumulative distribution function (CDF) relate to these calculations?

**Tip:** In a standard normal distribution, approximately 68% of data falls within 1 SD, 95% within 2 SDs, and 99.7% within 3 SDs of the mean (the empirical rule).

If bolt thread length is normally distributed, what is the probability that the thread length of a randomly selected bolt is (a) within 1.5 SDs of its mean value? (b) farther than 2.5 SDs from its mean value? (c) between 1 and 2 SDs from its mean value?

Find out how to calculate the probability of bolt thread length falling within 1.5, farther than 2.5, and between 1 and 2 standard deviations from the mean in a normally distributed dataset. Using key probability concepts, Z-scores, and the empirical rule, this solution is ideal for high school students in Grades 10-12.

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