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kirza4 [7]
3 years ago
5

Production of a gas flow meter takes place in two distinct operations. Measurements (n = 15) of the time required for the first

operation yield sample mean and standard deviation of 45 min and 4 min, respectively. Measurements (n = 15) of the time required for the second operation yield sample mean and standard deviation of 32 min and 2.5 min, respectively. For the purposes of this question, assume each time is a normally‐distributed random variable. a) What is the probability that the time required to complete both of those steps will exceed 80 min? b) What is a 95% confidence interval for the expected total time required to produce one flow meter? c) What is a 95%
Mathematics
1 answer:
Karolina [17]3 years ago
4 0

Answer:

a) P(a > 80) = 0.323

b) The 95% confidence interval = (73.40, 80.60)

c) The 95% confidence interval expresses that the mean of the distribution can always be found in the given range, with a 95% confidence level.

Step-by-step explanation:

X ~ (45, 4)

Y ~ (32, 2.5)

(X+Y) ~ (77, 6.5)

Let a = (X+Y)

a) Probability that the time required to complete both of those steps will exceed 80 min = P(a > 80)

This is a normal distribution problem

We then standardize 80 min time

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (a - μ)/σ = (80 - 77)/6.5 = 0.46

To determine the probability the time required to complete both of those steps will exceed 80 min = P(a > 80) = P(z > 0.46)

We'll use data from the normal probability table for these probabilities

P(a > 80) = P(z > 0.46) = 1 - P(z ≤ 0.46) = 1 - 0.677 = 0.323

b) 95% confidence interval for the expected total time required to produce one flow meter.

We need to obtain the margin of error

Margin of error = (critical value) × (standard error of the sample)

Critical value for a 95% confidence interval = critical value for a significance level of 5% = t(15-1, 0.05/2) = 2.145 (using the t-score since information on the population mean and standard deviation isn't known)

Standard error for the sample of sum of times = (standard deviation of the sum of times)/√n = (6.5/√15) = 1.678

Margin of error = 2.145 × 1.678 = 3.60

Limits of the confidence interval = (Sample mean ± margin of error)

Lower limit of the confidence interval = (Sample mean - margin of error) = 77 - 3.60 = 73.40

Upper limit of the confidence interval = 77 + 3.60 = 80.60

The confidence interval = (73.40, 80.60)

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