a. The mean of this distribution is b. The standard deviation is c. The probability that wave will crash onto the beach exactly 0.7 seconds after the person arrives is P(x 0.7) =
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- The mean amount of time it takes a kidney stone to pass is 12 days and the standard deviation is 5 days. Suppose that one individual is randomly chosen. Let X = time to pass the kidney stone. a. What is the distribution of X? X - N( b. Find the probability that a randomly selected person with a kidney stone will take longer than 8 days to pass it. Round to 4 decimal places.Suppose that the weight of an newborn fawn is Uniformly distributed between 2.1 and 3.8 kg. Suppose that a newborn fawn is randomly selected. Round answers to 4 decimal places when possible. a. The mean of this distribution is b. The standard deviation is c. The probability that fawn will weigh exactly 3.7 kg is P(x = 3.7) = d. The probability that a newborn fawn will be weigh between 2.6 and 2.8 is P(2.6 < x < 2.8) =Suppose that the weight of an newborn fawn is Uniformly distributed between 2.5 and 3.8 kg. Suppose that a newborn fawn is randomly selected. Round answers to 4 decimal places when possible.a. The mean of this distribution is b. The standard deviation is c. The probability that fawn will weigh exactly 3.7 kg is P(x = 3.7) = d. The probability that a newborn fawn will be weigh between 3.1 and 3.4 is P(3.1 < x < 3.4) = e. The probability that a newborn fawn will be weigh more than 3.36 is P(x > 3.36) = f. P(x > 2.6 | x < 3.3) = g. Find the 58th percentile. solve e,f and g only.
- Assume that adults have IQ scores that are normally distributed with a mean of 95.3 and a standard deviation of 16.4. Find the probability that a randomly selected adult has an IQ greater than 125.0. (Hint: Draw a graph.) The probability that a randomly selected adult from this group has an IQ greater than 125.0 is ☐ . (Round to four decimal places as needed.)Suppose that the weight of an newborn fawn is Uniformly distributed between 2.2 and 3.9 kg. Suppose that a newborn fawn is randomly selected. Round answers to 4 decimal places when possible. The mean of this distribution is . The standard deviation is . The probability that fawn will weigh exactly 2.5 kg is P(x=2.5)=P(x=2.5)= . The probability that a newborn fawn will be weigh between 2.6 kg and 3.7 kg is P(2.6<x<3.7)=P(2.6<x<3.7)= . The probability that a newborn fawn will be weigh more than 3.24 kg is P(x>3.24)=P(x>3.24)= . P(x>2.8∣x<3.3)=P(x>2.8∣x<3.3)= . Find the 45th percentile. kgToday, the waves are crashing onto the beach every 5.3 seconds. The times from when a person arrives at the shoreline until a crashing wave is observed follows a Uniform distribution from 0 to 5.3 seconds. Round to 4 decimal places where possible. a. The mean of this distribution is b. The standard deviation is c. The probability that wave will crash onto the beach exactly 3.3 seconds after the person arrives is P(x = 3.3) = d. The probability that the wave will crash onto the beach between 1.9 and 2.3 seconds after the person arrives is P(1.9 3.26) = f. Find the minimum for the upper quartile. seconds.
- Today, the waves are crashing onto the beach every 4.5 seconds. The times from when a person arrives at the shoreline until a crashing wave is observed follows a Uniform distribution from 0 to 4.5 seconds. Round to 4 decimal places where possible. a. The mean of this distribution is b. The standard deviation is c. The probability that wave will crash onto the beach exactly 1.7 seconds after the person arrives is P(x = 1.7) = d. The probability that the wave will crash onto the beach between 1.5 and 3.9 seconds after the person arrives is P(1.5 1) = f. Find the maximum for the lower quartile. seconds.Solve for the mean and standard deviation of the following Binomial distributions. a. n = 20 and p = 0.70 b. n = 70 and p = 0.35 c. n = 100 and p = 0.50Today, the waves are crashing onto the beach every 4.1 seconds. The times from when a person arrives at the shoreline until a crashing wave is observed follows a Uniform distribution from 0 to 4.1 seconds. Round to 4 decimal places where possible. a. The mean of this distribution is b. The standard deviation is C. The probability that wave will crash onto the beach exactly 2.8 seconds after the person arrives is P(x = 2.8) = d. The probability that the wave will crash onto the beach between 0.4 and 1.9 seconds after the person arrives is P(0.4 0.82) = f. Suppose that the person has already been standing at the shoreline for 0.5 seconds without a wave crashing in. Find the probability that it will take between 1 and 4 seconds for the wave to crash onto the shoreline. g. 26% of the time a person will wait at least how long before the wave crashes in? seconds. h. Find the maximum for the lower quartile. seconds.
- Today, the waves are crashing onto the beach every 5.4 seconds. The times from when a person arrives at the shoreline until a crashing wave is observed follows a Uniform distribution from 0 to 5.4 seconds. Round to 4 decimal places where possible. a. The mean of this distribution is 2.7 b. The standard deviation is c. The probability that wave will crash onto the beach exactly 4.1 seconds after the person arrives is P(x = 4.1) =| d. The probability that the wave will crash onto the beach between 1.8 and 4.8 seconds after the person arrives is P(1.8 2.28) = f. Find the maximum for the lower quartile. seconds.Today, the waves are crashing onto the beach every 4.2 seconds. The times from when a person arrives at the shoreline until a crashing wave is observed follows a Uniform distribution from 0 to 4.2 seconds. Round to 4 decimal places where possible. a. The mean of this distribution is b. The standard deviation is C. The probability that it will take longer than 2.14 seconds for the wave to crash onto the beach after the person arrives is P(x > 2.14) = %3DOn average, indoor cats live to 14 years old with a standard deviation of 2.7 years. Suppose that the distribution is normal. Let X = the age at death of a randomly selected indoor cat. Round answers to 4 decimal places where possible.A. The middle 50% of indoor cats' age of death lies between what two numbers? Low: years High: years