Hawkes Chapter 8 Principles of Statistics

The annual increase in height of cedar trees is believed to be distributed uniformly between six and eleven inches. Find the probability that a randomly selected cedar tree will grow less than 9 inches in a given year. Round your answer to four decimal places, if necessary.

Let X= the number of inches a randomly selected cedar tree grows in a given year. The probability of a random variable is given by the area under the probability density function. For the uniform distribution, the probability is simply calculated using the formula for the area of a rectangle. Area=Width⋅Height According to the uniform probability density function f(x)=1/b−a. Therefore, Height = 1/b−a = 1/11−6 = 1/5. We are interested in the probability that a cedar tree will grow less than 9 inches, so Width = 9−6 = 3 and Area = Width⋅Height = 3(1/5) = 0.6. Since the probability of observing a uniform random variable is the area under the density function associated with the interval, we calculate the probability that a cedar tree will grow less than 9 inches in a given year as follows. P(X<9) = 0.6

The annual increase in height of cedar trees is believed to be distributed uniformly between five and twelve inches. Find the probability that a randomly selected cedar tree will grow less than 6 inches in a given year. Round your answer to four decimal places, if necessary.

Let X= the number of inches a randomly selected cedar tree grows in a given year. The probability of a random variable is given by the area under the probability density function. For the uniform distribution, the probability is simply calculated using the formula for the area of a rectangle. Area=Width⋅Height According to the uniform probability density function f(x)=1/b−a. Therefore, Height = 1/b−a = 1/12−5 = 1/7. We are interested in the probability that a cedar tree will grow less than 6 inches, so Width = 7−6 = 1 and Area = Width⋅Height = 1(1/7) = 0.1429. Since the probability of observing a uniform random variable is the area under the density function associated with the interval, we calculate the probability that a cedar tree will grow less than 6 inches in a given year as follows. P(X<6) = 0.1429

The annual increase in height of cedar trees is believed to be distributed uniformly between five and twelve inches. Find the probability that a randomly selected cedar tree will grow less than 7 inches in a given year. Round your answer to four decimal places, if necessary.

Let X= the number of inches a randomly selected cedar tree grows in a given year. The probability of a random variable is given by the area under the probability density function. For the uniform distribution, the probability is simply calculated using the formula for the area of a rectangle. Area=Width⋅Height According to the uniform probability density function f(x)=1/b−a. Therefore, Height = 1/b−a = 1/12−5 = 1/7. We are interested in the probability that a cedar tree will grow less than 7 inches, so Width = 7−5 = 2 and Area = Width⋅Height = 2(1/7) = 0.2857. Since the probability of observing a uniform random variable is the area under the density function associated with the interval, we calculate the probability that a cedar tree will grow less than 7 inches in a given year as follows. P(X>7) = 0.2857

The annual increase in height of cedar trees is believed to be distributed uniformly between seven and twelve inches. Find the probability that a randomly selected cedar tree will grow more than 11 inches in a given year. Round your answer to four decimal places, if necessary.

Let X= the number of inches a randomly selected cedar tree grows in a given year. The probability of a random variable is given by the area under the probability density function. For the uniform distribution, the probability is simply calculated using the formula for the area of a rectangle. Area=Width⋅Height According to the uniform probability density function f(x)=1/b−a. Therefore, Height = 1/b−a = 1/12−7 = 1/5. We are interested in the probability that a cedar tree will grow less than 11 inches, so Width = 12−11 = 1 and Area = Width⋅Height = 1(1/5) = 0.2. Since the probability of observing a uniform random variable is the area under the density function associated with the interval, we calculate the probability that a cedar tree will grow less than 11 inches in a given year as follows. P(X<11) = 0.2

The annual increase in height of cedar trees is believed to be distributed uniformly between nine and fourteen inches. Find the probability that a randomly selected cedar tree will grow more than 12 inches in a given year. Round your answer to four decimal places, if necessary.

Let X= the number of inches a randomly selected cedar tree grows in a given year. The probability of a random variable is given by the area under the probability density function. For the uniform distribution, the probability is simply calculated using the formula for the area of a rectangle. Area=Width⋅Height According to the uniform probability density function f(x)=1/b−a. Therefore, Height = 1/b−a = 1/14−9 = 1/5. We are interested in the probability that a cedar tree will grow less than 6 inches, so Width = 14−12 = 2 and Area = Width⋅Height = 2(1/5) = 0.4. Since the probability of observing a uniform random variable is the area under the density function associated with the interval, we calculate the probability that a cedar tree will grow less than 6 inches in a given year as follows. P(X>11) = 0.4

A particular employee arrives at work sometime between 8:00 a.m. and 8:30 a.m. Based on past experience the company has determined that the employee is equally likely to arrive at any time between 8:00 a.m. and 8:30 a.m. Find the probability that the employee will arrive between 8:15 a.m. and 8:20 a.m. Round your answer to four decimal places, if necessary.

Let X=the time when the employee arrives at work. The probability of a random variable is given by the area under the probability density function. For the uniform distribution, the probability is simply calculated using the formula for the area of a rectangle. Area = Width⋅Height According to the uniform probability density function f(x)=1/b−a. Therefore, Height = 1/b−a = 1/30−0 = 1/30. We are interested in the probability that the employee will arrive between 8:15 a.m. and 8:20 a.m., so Width = 20−15 = 5, and Area= Width⋅Height = 5(1/30) = 0.1667 Since the probability of observing a uniform random variable is the area under the density function associated with the interval, we calculate the probability that the employee will arrive between 8:15 a.m. and 8:20 a.m. as follows. P(8:15≤X≤8:20) = 0.1667

A particular employee arrives at work sometime between 8:00 a.m. and 8:30 a.m. Based on past experience the company has determined that the employee is equally likely to arrive at any time between 8:00 a.m. and 8:30 a.m. Find the probability that the employee will arrive between 8:15 a.m. and 8:25 a.m. Round your answer to four decimal places, if necessary.

Let X=the time when the employee arrives at work. The probability of a random variable is given by the area under the probability density function. For the uniform distribution, the probability is simply calculated using the formula for the area of a rectangle. Area = Width⋅Height According to the uniform probability density function f(x)=1/b−a. Therefore, Height = 1/b−a = 1/30−0 = 1/30. We are interested in the probability that the employee will arrive between 8:15 a.m. and 8:25 a.m., so Width= 25−15=10, and Area= Width⋅Height = 10(1/30) = 0.3333 Since the probability of observing a uniform random variable is the area under the density function associated with the interval, we calculate the probability that the employee will arrive between 8:15 a.m. and 8:25 a.m. as follows. P(8:15≤X≤8:25) = 0.3

A particular employee arrives at work sometime between 8:00 a.m. and 8:40 a.m. Based on past experience the company has determined that the employee is equally likely to arrive at any time between 8:00 a.m. and 8:40 a.m. Find the probability that the employee will arrive between 8:15 a.m. and 8:20 a.m. Round your answer to four decimal places, if necessary.

Let X=the time when the employee arrives at work. The probability of a random variable is given by the area under the probability density function. For the uniform distribution, the probability is simply calculated using the formula for the area of a rectangle. Area = Width⋅Height According to the uniform probability density function f(x)=1/b−a. Therefore, Height = 1/b−a = 1/40−0 = 1/40. We are interested in the probability that the employee will arrive between 8:15 a.m. and 8:20 a.m., so Width = 20−15 = 5, and Area= Width⋅Height = 5(1/40) = 0.125 Since the probability of observing a uniform random variable is the area under the density function associated with the interval, we calculate the probability that the employee will arrive between 8:15 a.m. and 8:20 a.m. as follows. P(8:15≤X≤8:20) = 0.125

A particular employee arrives at work sometime between 8:00 a.m. and 8:50 a.m. Based on past experience the company has determined that the employee is equally likely to arrive at any time between 8:00 a.m. and 8:50 a.m. Find the probability that the employee will arrive between 8:30 a.m. and 8:45 a.m. Round your answer to four decimal places, if necessary.

Let X=the time when the employee arrives at work. The probability of a random variable is given by the area under the probability density function. For the uniform distribution, the probability is simply calculated using the formula for the area of a rectangle. Area = Width⋅Height According to the uniform probability density function f(x)=1/b−a. Therefore, Height = 1/b−a = 1/50−0 = 1/50. We are interested in the probability that the employee will arrive between 8:30 a.m. and 8:45 a.m., so Width = 45−30 = 15, and Area= Width⋅Height = 15(1/50) = 0.3 Since the probability of observing a uniform random variable is the area under the density function associated with the interval, we calculate the probability that the employee will arrive between 8:30 a.m. and 8:45 a.m. as follows. P(8:30≤X≤8:45) = 0.3

Polar Bear Frozen Foods manufactures frozen French fries for sale to grocery store chains. The final package weight is thought to be a uniformly distributed random variable. Assume X, the weight of French fries, has a uniform distribution between 52 ounces and 56 ounces. What is the mean weight for a package? What is the standard deviation for the weight of a package? Round your answers to four decimal places, if necessary.

The mean of a continuous uniform random variable is given by μ=a+b/2 = 52+56/2=54. Therefore, the mean weight for a package is 62 ounces. The standard deviation of a continuous uniform random variable is given by σ=b−a/√12 = 56−52/√12≈2.3094. Therefore, the standard deviation for the weight of a package is 1.1547 ounces.

Polar Bear Frozen Foods manufactures frozen French fries for sale to grocery store chains. The final package weight is thought to be a uniformly distributed random variable. Assume X, the weight of French fries, has a uniform distribution between 57 ounces and 65 ounces. What is the mean weight for a package? What is the standard deviation for the weight of a package? Round your answers to four decimal places, if necessary.

The mean of a continuous uniform random variable is given by μ=a+b/2 = 57+65/2=61. Therefore, the mean weight for a package is 62 ounces. The standard deviation of a continuous uniform random variable is given by σ=b−a/√12 = 65−57/√12≈2.3094. Therefore, the standard deviation for the weight of a package is 2.3094 ounces.

Polar Bear Frozen Foods manufactures frozen French fries for sale to grocery store chains. The final package weight is thought to be a uniformly distributed random variable. Assume X, the weight of French fries, has a uniform distribution between 50 ounces and 60 ounces. What is the mean weight for a package? What is the standard deviation for the weight of a package? Round your answers to four decimal places, if necessary.

The mean of a continuous uniform random variable is given by μ=a+b/2=50+60/2=55. Therefore, the mean weight for a package is 55 ounces. The standard deviation of a continuous uniform random variable is given by σ=b−a/√12=60−50/√12≈2.8868. Therefore, the standard deviation for the weight of a package is 2.8868 ounces

Polar Bear Frozen Foods manufactures frozen French fries for sale to grocery store chains. The final package weight is thought to be a uniformly distributed random variable. Assume X, the weight of French fries, has a uniform distribution between 58 ounces and 66 ounces. What is the mean weight for a package? What is the standard deviation for the weight of a package? Round your answers to four decimal places, if necessary.

The mean of a continuous uniform random variable is given by μ=a+b/2=58+66/2=62. Therefore, the mean weight for a package is 62 ounces. The standard deviation of a continuous uniform random variable is given by σ=b−a/√12=66−58/√12≈2.3094. Therefore, the standard deviation for the weight of a package is 2.3094 ounces.