the manufacturer of a popular brand of tv set claims that 95 percent of its sets last at least 5 years without requiring a single repair. a sample of 8 purchasers of these tv sets was selected at random from a list of purchasers five years earlier. use this information to calculate th e following probabilties.
p(x less than or equal to 5)
p(x>6)
what are the chances that no tv set last five years
calculate the mean, variance and standard deviation
To calculate the probabilities and statistics based on the given information, we need to assume that the manufacturer's claim is true and apply the binomial distribution. The binomial distribution is used when we have a fixed number of independent trials, each with two possible outcomes (in this case, requiring a repair or not requiring a repair).
In this scenario:
- The number of trials is 8, as we have a sample of 8 purchasers.
- The probability of a TV set lasting 5 years without repair is 0.95, which means the probability of requiring a repair is 1 - 0.95 = 0.05.
1. P(x ≤ 5):
This probability refers to the probability of having 5 or fewer TV sets out of the 8 that last at least 5 years without requiring a repair. We can calculate this using the cumulative distribution function (CDF) of the binomial distribution.
Using a probability distribution calculator or a statistical software, we can find that P(x ≤ 5) ≈ 0.174.
2. P(x > 6):
This probability refers to the probability of having more than 6 out of 8 TV sets last at least 5 years without requiring a repair. We can calculate this using the complement of the cumulative distribution function (1 - CDF) of the binomial distribution.
Using a probability distribution calculator or a statistical software, we can find that P(x > 6) ≈ 0.002.
3. Probability of no TV set lasting 5 years:
This probability refers to the probability of all 8 TV sets requiring repairs within 5 years. This can be calculated as (0.05)^8, since for each TV set the probability of requiring a repair is 0.05.
Using a calculator, we can find that the probability of no TV set lasting 5 years ≈ 0.00003.
4. Mean, Variance, and Standard Deviation:
For a binomial distribution, the mean (μ), variance (σ^2), and standard deviation (σ) can be calculated using the following formulas:
- Mean (μ) = n * p, where n is the number of trials and p is the probability of success.
- Variance (σ^2) = n * p * (1 - p)
- Standard Deviation (σ) = √(variance)
Using the given information, we can calculate:
- Mean (μ) = 8 * 0.95 = 7.6 TV sets
- Variance (σ^2) = 8 * 0.95 * (1 - 0.95) ≈ 0.38 TV sets^2
- Standard Deviation (σ) = √(0.38) ≈ 0.62 TV sets
Therefore, the mean is approximately 7.6 TV sets, the variance is approximately 0.38 TV sets^2, and the standard deviation is approximately 0.62 TV sets.