For a certain type of computers, the length of time between charges of the battery is normally distributed with a mean of 50hours and a standard deviation of 13 hours. John owns one of these computers.29. How many standard deviations away from the mean is a battery that last 30 hours?30. What is the probability that his will charge will last between 50 and 70 hours?31. What is the probability that his charge will last less than 40 hours?32. What is the probability that his charge will last more than 70 hours?33. What is the cut off for an unusually short time for a charge to last?34. What is the cut off for an unusually long time for a charge to last?decdecdec-1.538%%%unitsunits

The objective of the question is to understand the properties of a normal distribution and apply them to a real-world scenario. In this case, we are given a mean (μ) of 50 hours and a standard deviation (σ) of 13 hours for the battery life of a certain type of computer. We are asked to calculate various probabilities and z-scores related to this distribution.29. To find out how many standard deviations away from the mean a battery that lasts 30 hours is, we need to calculate the z-score. The z-score is calculated as (X - μ) / σ, where X is the value we are interested in, μ is the mean, and σ is the standard deviation. Substituting the given values, we get (30 - 50) / 13 = -20 / 13 = -1.54. So, a battery that lasts 30 hours is approximately 1.54 standard deviations below the mean.30. To find the probability that the charge will last between 50 and 70 hours, we need to calculate the z-scores for these values and look up the corresponding probabilities in the standard normal distribution table. The z-score for 50 hours is (50 - 50) / 13 = 0, and for 70 hours it is (70 - 50) / 13 = 20 / 13 = 1.54. The probability for a z-score of 0 is 0.5 (or 50%), and for a z-score of 1.54 is approximately 0.9382. The probability that the charge will last between 50 and 70 hours is therefore 0.9382 - 0.5 = 0.4382, or approximately 43.82%.31. To find the probability that the charge will last less than 40 hours, we calculate the z-score for 40 hours, which is (40 - 50) / 13 = -10 / 13 = -0.77. The probability for a z-score of -0.77 is approximately 0.2206, or 22.06%. So, the probability that the charge will last less than 40 hours is approximately 22.06%.32. To find the probability that the charge will last more than 70 hours, we calculate the z-score for 70 hours, which is (70 - 50) / 13 = 20 / 13 = 1.54. The probability for a z-score of 1.54 is approximately 0.9382. However, since we are interested in the probability that the charge will last more than 70 hours, we need to subtract this probability from 1. So, the probability that the charge will last more than 70 hours is 1 - 0.9382 = 0.0618, or approximately 6.18%.33. An unusually short time for a charge to last is typically defined as a time that is more than two standard deviations below the mean. This is calculated as μ - 2σ. Substituting the given values, we get 50 - 2*13 = 50 - 26 = 24 hours. So, any time less than 24 hours would be considered unusually short for a charge to last.34. Similarly, an unusually long time for a charge to last is typically defined as a time that is more than two standard deviations above the mean. This is calculated as μ + 2σ. Substituting the given values, we get 50 + 2*13 = 50 + 26 = 76 hours. So, any time more than 76 hours would be considered unusually long for a charge to last.In summary, a battery that lasts 30 hours is approximately 1.54 standard deviations below the mean. The probability that the charge will last between 50 and 70 hours is approximately 43.82%, the probability that it will last less than 40 hours is approximately 22.06%, and the probability that it will last more than 70 hours is approximately 6.18%. An unusually short time for a charge to last is less than 24 hours, and an unusually long time is more than 76 hours.