To answer these questions, we need to convert the given times into z-scores and then use the standard normal distribution table to find the probabilities.

(a) To find the probability that a call lasted less than 210 seconds, we first convert 210 seconds into a z-score using the formula z = (X - μ) / σ, where X is the value we're interested in, μ is the mean, and σ is the standard deviation. So, z = (210 - 300) / 50 = -1.8. Looking up -1.8 in the z-table gives us a probability of 0.0359, or 3.59%.

(b) To find the probability that a call lasted between 210 and 320 seconds, we convert both times into z-scores. For 210 seconds, we already found that z = -1.8. For 320 seconds, z = (320 - 300) / 50 = 0.4. Looking up 0.4 in the z-table gives us a probability of 0.6554. To find the probability that a call lasted between 210 and 320 seconds, we subtract the probability of a call lasting less than 210 seconds from the probability of a call lasting less than 320 seconds: 0.6554 - 0.0359 = 0.6195, or 61.95%.

(c) To find the probability that a call lasted between 100 and 210 seconds, we convert 100 seconds into a z-score: z = (100 - 300) / 50 = -4. Looking up -4 in the z-table gives us a probability of nearly 0. Subtracting this from the probability of a call lasting less than 210 seconds gives us 0.0359, or 3.59%.

(d) To find the time below which 10% of all calls will last, we look up the z-score that corresponds to a cumulative probability of 0.10 in the z-table, which is -1.28. We then convert this z-score back into a time using the formula X = μ + zσ, where z is the z-score, μ is the mean, and σ is the standard deviation. So, X = 300 + (-1.28)(50) = 236 seconds. So, 10% of all calls will last less than 236 seconds.

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To answer these questions, we need to convert the given times into z-scores and then use the standard normal distribution table to find the probabilities.

(a) To find the probability that a call lasted less than 210 seconds, we first convert 210 seconds into a z-score using the formula z = (X - μ) / σ, where X is the value we're interested in, μ is the mean, and σ is the standard deviation. So, z = (210 - 300) / 50 = -1.8. Looking up -1.8 in the z-table gives us a probability of 0.0359, or 3.59%.

(b) To find the probability that a call lasted between 210 and 320 seconds, we convert both times into z-scores. For 210 seconds, we already found that z = -1.8. For 320 seconds, z = (320 - 300) / 50 = 0.4. Looking up 0.4 in the z-table gives us a probability of 0.6554. To find the probability that a call lasted between 210 and 320 seconds, we subtract the probability of a call lasting less than 210 seconds from the probability of a call lasting less than 320 seconds: 0.6554 - 0.0359 = 0.6195, or 61.95%.

(c) To find the probability that a call lasted between 100 and 210 seconds, we convert 100 seconds into a z-score: z = (100 - 300) / 50 = -4. Looking up -4 in the z-table gives us a probability of nearly 0. Subtracting this from the probability of a call lasting less than 210 seconds gives us 0.0359, or 3.59%.

(d) To find the time below which 10% of all calls will last, we look up the z-score that corresponds to a cumulative probability of 0.10 in the z-table, which is -1.28. We then convert this z-score back into a time using the formula X = μ + zσ, where z is the z-score, μ is the mean, and σ is the standard deviation. So, X = 300 + (-1.28)(50) = 236 seconds. So, 10% of all calls will last less than 236 seconds.

Knowee AI · Accepted Answer

To answer these questions, we need to convert the given times into z-scores and then use the standard normal distribution table to find the probabilities.

(a) To find the probability that a call lasted less than 210 seconds, we first convert 210 seconds into a z-score using the formula z = (X - μ) / σ, where X is the value we're interested in, μ is the mean, and σ is the standard deviation. So, z = (210 - 300) / 50 = -1.8. Looking up -1.8 in the z-table gives us a probability of 0.0359, or 3.59%.

(b) To find the probability that a call lasted between 210 and 320 seconds, we convert both times into z-scores. For 210 seconds, we already found that z = -1.8. For 320 seconds, z = (320 - 300) / 50 = 0.4. Looking up 0.4 in the z-table gives us a probability of 0.6554. To find the probability that a call lasted between 210 and 320 seconds, we subtract the probability of a call lasting less than 210 seconds from the probability of a call lasting less than 320 seconds: 0.6554 - 0.0359 = 0.6195, or 61.95%.

(c) To find the probability that a call lasted between 100 and 210 seconds, we convert 100 seconds into a z-score: z = (100 - 300) / 50 = -4. Looking up -4 in the z-table gives us a probability of nearly 0. Subtracting this from the probability of a call lasting less than 210 seconds gives us 0.0359, or 3.59%.

(d) To find the time below which 10% of all calls will last, we look up the z-score that corresponds to a cumulative probability of 0.10 in the z-table, which is -1.28. We then convert this z-score back into a time using the formula X = μ + zσ, where z is the z-score, μ is the mean, and σ is the standard deviation. So, X = 300 + (-1.28)(50) = 236 seconds. So, 10% of all calls will last less than 236 seconds.

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