Math  /  Data & Statistics

QuestionA manufacturer knows that their items have a normally distributed lifespan, with a mean of 12.4 years, and standard deviation of 3.9 years.
If you randomly purchase one item, what is the probability it will last longer than 20 years? Please show your answer to 4 decimal places.
Probability == \square

Studdy Solution

STEP 1

1. The lifespan of the items follows a normal distribution with a mean (μ\mu) of 12.4 years and a standard deviation (σ\sigma) of 3.9 years.
2. We need to find the probability that the lifespan of a randomly selected item exceeds 20 years.
3. This problem can be solved using the properties of the normal distribution and the standard normal distribution (z-scores).

STEP 2

1. Convert the raw lifespan value (20 years) to a z-score.
2. Use the z-score to find the corresponding probability from the standard normal distribution.
3. Subtract the found probability from 1 to get the probability of the item lasting longer than 20 years.

STEP 3

Calculate the z-score for the lifespan of 20 years using the formula for z-score: z=Xμσ z = \frac{X - \mu}{\sigma} where X X is the value of interest (20 years), μ \mu is the mean (12.4 years), and σ \sigma is the standard deviation (3.9 years).
z=2012.43.9 z = \frac{20 - 12.4}{3.9} z=7.63.9 z = \frac{7.6}{3.9} z1.9487 z \approx 1.9487

STEP 4

Find the cumulative probability corresponding to the z-score of 1.9487 using standard normal distribution tables or a calculator that provides cumulative distribution function (CDF) values for the normal distribution.
P(Z1.9487)0.9744 P(Z \leq 1.9487) \approx 0.9744

STEP 5

Calculate the probability that the item lasts longer than 20 years by subtracting the cumulative probability from 1.
P(X>20)=1P(Z1.9487) P(X > 20) = 1 - P(Z \leq 1.9487) P(X>20)=10.9744 P(X > 20) = 1 - 0.9744 P(X>20)0.0256 P(X > 20) \approx 0.0256
Therefore, the probability that a randomly selected item will last longer than 20 years is approximately 0.0256.
Probability == 0.02560.0256

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