Assume That X Is Normally Distributed With Mean 6

Assume That X Is Normally Distributed With A Mean Of 6 And A Standard

Determine the value for x for each of the following, given that X is normally distributed with a mean of 6 and a standard deviation of 3:

  • (a) P(X > x) = 0.95
  • (b) P(x
  • (c) P(3

The tensile strength of paper follows a normal distribution with a mean of 35 pounds per square inch (psi) and a standard deviation of 2 psi. Calculate:

  • (a) The probability that a sample's tensile strength is less than 39 psi.
  • (b) The proportion of samples scrapped if the strength must exceed 29 psi.

Suppose that X represents time measurements from a gamma distribution with a mean of 4 minutes and a variance of 2 minutes2. Find the parameters λ and r.

Paper For Above instruction

In this paper, we will address three statistical problems involving normal and gamma distributions based on the given parameters and conditions. The first part involves calculating specific quantiles and probabilities for a normal distribution with a known mean and standard deviation. The second part uses the properties of the normal distribution to find probabilities related to tensile strength measurements, and the third involves determining the parameters of a gamma distribution given its mean and variance.

Part 1: Normal Distribution with Mean 6 and Standard Deviation 3

Given a normally distributed random variable X with mean μ = 6 and standard deviation σ = 3, we need to find specific values of x based on probability statements.

(a) Finding x such that P(X > x) = 0.95

This probability indicates that the value x is at the 5th percentile of the distribution. Since the total probability is 1, the left tail area up to x is 0.05.

To find x, we first identify the z-score corresponding to a cumulative probability of 0.05 using standard normal distribution tables or statistical software. The z-score for P(Z

Next, convert this z-score to the original distribution using the formula:

x = μ + zσ = 6 + (-1.645)(3) ≈ 6 - 4.935 ≈ 1.065

Therefore, the value of x is approximately 1.07 pounds per square inch.

(b) Finding x such that P(x

We interpret this as the probability that X lies between some x and 9, totaling 0.2. To determine x, we need the cumulative probabilities corresponding to the bounds.

Since P(X

Calculate the z-score for X = 9:

z = (9 - 6) / 3 = 3 / 3 = 1

The cumulative probability P(Z

P(Z

Next, find z_x such as P(Z

Now, find x:

x = μ + z_xσ = 6 + 0.37 * 3 ≈ 6 + 1.11 ≈ 7.11

Thus, the value of x is approximately 7.11 pounds per square inch.

(c) Finding x such that P(3

This implies that the probability that X falls between 3 and x is 0.8. We first find the cumulative probability at X = 3:

Calculate z for X = 3:

z = (3 - 6) / 3 = -3 / 3 = -1

The cumulative probability P(Z

P(X 3) + P(3

Find z corresponding to P(Z

Calculate x:

x = μ + zσ = 6 + 1.75 * 3 = 6 + 5.25 = 11.25

Hence, the value of x is approximately 11.25 pounds per square inch.

Part 2: Tensile Strength of Paper

The tensile strength follows a normal distribution with mean μ = 35 psi and standard deviation σ = 2 psi.

(a) Probability that a sample is less than 39 psi

Calculate the z-score:

z = (39 - 35) / 2 = 4 / 2 = 2

The cumulative probability P(Z

Thus, the probability that a randomly selected sample has tensile strength less than 39 psi is approximately 97.72%.

(b) Proportion of samples scrapped if strength must exceed 29 psi

Calculate the z-score for 29 psi:

z = (29 - 35) / 2 = -6 / 2 = -3

The probability P(Z

Since scrap is produced when the tensile strength is below 29 psi, nearly 0.13% of samples are scrapped under this criterion.

Part 3: Gamma Distribution Parameters from Mean and Variance

Given that X is gamma-distributed with mean μ = 4 minutes and variance σ2 = 2 minutes2, we find its shape (r) and rate (λ) parameters.

Parameter r (shape)

The mean and variance of a gamma distribution are related to r and λ by:

  • μ = r / λ
  • σ2 = r / λ2

Dividing the variance by the mean:

σ2 / μ = (r / λ2 ) / (r / λ) = λ

Computing λ:

λ = σ2 / μ = 2 / 4 = 0.5

Now, solving for r:

r = μ λ = 4 0.5 = 2

Final gamma parameters:

  • Shape parameter, r = 2
  • Rate parameter, λ = 0.5

Conclusion

This analysis demonstrates the application of normal and gamma distributions to practical problems involving probabilistic and statistical calculations. The solutions obtained serve as foundational tools for quality control, reliability assessment, and statistical inference in engineering and manufacturing processes, underpinning the importance of understanding distribution parameters and their applications.

References

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