unit data and statistics homework 6 answer key

Uniform Distribution

Normal Distribution

P (6 < x < 7)

The probability is equal to the area from x = 3 2 3 2 to x = 4 above the x-axis and up to f ( x ) = 1 3 1 3 .

It means that the value of x is just as likely to be any number between 1.5 and 4.5.

1.5 ≤ x ≤ 4.5

b is 12, and it represents the highest value of x .

X = The age (in years) of cars in the staff parking lot

f ( x ) = 1 9 1 9 where x is between 0.5 and 9.5, inclusive.

• Check student’s solution.
• 3.5 7 3.5 7
• Check student's solution.

No, outcomes are not equally likely. In this distribution, more people require a little bit of time, and fewer people require a lot of time, so it is more likely that someone will require less time.

f ( x ) = 0.2e -0.2 x

f ( x ) = 0.75 e -0.75 x

The mean is larger. The mean is 1 m = 1 0.75 ≈ 1.33 1 m = 1 0.75 ≈ 1.33 , which is greater than 0.9242.

m = 0.000121

• Check student's solution
• P ( x < 5,730) = 0.5001
• k = 2947.73

Age is a measurement, regardless of the accuracy used.

• X ~ U (1, 9)
• f ( x ) = 1 8 f ( x ) = 1 8 where 1 ≤ x ≤ 9 1 ≤ x ≤ 9
• 15 32 15 32
• 333 800 333 800
• X represents the length of time a commuter must wait for a train to arrive on the Red Line.
• X ~ U (0, 8)
• Graph the probability distribution.
• f ( x ) = 1 8 f ( x ) = 1 8 where 0 ≤ x ≤ 8 0 ≤ x ≤ 8
• The area must be 0.25, and 0.25 = (width) ( 1 9 ) ( 1 9 ) , so width = (0.25)(9) = 2.25. Thus, the value is 25 – 2.25 = 22.75.
• Draw the graph where a is now 18 and b is still 25. The height is 1 ( 25 − 18 ) 1 ( 25 − 18 ) = 1 7 1 7 So, P ( x > 21| x > 18) = (25 – 21) ( 1 7 ) ( 1 7 ) = 4/7.
• Use the formula: P ( x > 21| x > 18) = P ( x > 21  AND  x > 18 ) P ( x > 18 ) P ( x > 21  AND  x > 18 ) P ( x > 18 ) = P ( x > 21 ) P ( x > 18 ) P ( x > 21 ) P ( x > 18 ) = ( 25 − 21 ) ( 25 − 18 ) ( 25 − 21 ) ( 25 − 18 ) = 4 7 4 7 .
• P ( X > 650) = 700 − 650 700 − 300 = 50 400 = 1 8 700 − 650 700 − 300 = 50 400 = 1 8 = 0.125
• P (400 < X < 650) = 650 − 400 700 − 300 = 250 400 650 − 400 700 − 300 = 250 400 = 0.625
• 0.10 = width 700 − 300 width 700 − 300 , so width = 400(0.10) = 40. Since 700 – 40 = 660, the drivers travel at least 660 miles on the furthest 10% of days.
• X = the useful life of a particular car battery, measured in months.
• X is continuous.
• X ~ Exp (0.025)
• X = the time (in years) after reaching age 60 that it takes an individual to retire
• X ~ Exp ( 1 5 ) ( 1 5 )

Let T = the life time of a light bulb.

The decay parameter is m = 1/8, and T ∼ Exp(1/8). The cumulative distribution function is P ( T < t ) = 1 − e − t 8 P ( T < t ) = 1 − e − t 8

• Therefore, P ( T < 1) = 1 – e – 1 8 – 1 8 ≈ 0.1175.
• We want to find 0.02 = P ( T < t ) = 1 – e – t 8 – t 8 . Solving for t , e – t 8 – t 8 = 0.98, so – t 8 – t 8 = ln (0.98), and t = –8 ln (0.98) ≈ 0.1616 years, or roughly two months. The warranty should cover light bulbs that last less than 2 months. Or use ln (area_to_the_right) ( – m) = ln (1 – 0.2 ) – 1 8 ln (area_to_the_right) ( – m) = ln (1 – 0.2 ) – 1 8 = 0.1616.
• We must find P ( T < 8| T > 7). Notice that by the rule of complement events, P ( T < 8| T > 7) = 1 – P ( T > 8| T > 7). By the memoryless property ( P ( X > r + t | X > r ) = P ( X > t )). So P ( T > 8| T > 7) = P ( T > 1) = 1 – ( 1 – e – 1 8 ) = e – 1 8 ≈ 0.8825 1 – ( 1 – e – 1 8 ) = e – 1 8 ≈ 0.8825 Therefore, P ( T < 8| T > 7) = 1 – 0.8825 = 0.1175.

Let X = the number of no-hitters throughout a season. Since the duration of time between no-hitters is exponential, the number of no-hitters per season is Poisson with mean λ = 3. Therefore, ( X = 0) = 3 0 e – 3 0 ! 3 0 e – 3 0 ! = e –3 ≈ 0.0498

You could let T = duration of time between no-hitters. Since the time is exponential and there are 3 no-hitters per season, then the time between no-hitters is 1 3 1 3 season. For the exponential, µ = 1 3 1 3 . Therefore, m = 1 μ 1 μ = 3 and T ∼ Exp (3).

• The desired probability is P ( T > 1) = 1 – P ( T < 1) = 1 – (1 – e –3 ) = e –3 ≈ 0.0498.
• Let T = duration of time between no-hitters. We find P ( T > 2| T > 1), and by the memoryless property this is simply P ( T > 1), which we found to be 0.0498 in part a.
• Let X = the number of no-hitters is a season. Assume that X is Poisson with mean λ = 3. Then P ( X > 3) = 1 – P ( X ≤ 3) = 0.3528.
• 100 9 100 9 = 11.11
• P ( X > 10) = 1 – P ( X ≤ 10) = 1 – Poissoncdf(11.11, 10) ≈ 0.5532.
• The number of people with Type B blood encountered roughly follows the Poisson distribution, so the number of people X who arrive between successive Type B arrivals is roughly exponential with mean μ = 9 and m = 1 9 1 9 . The cumulative distribution function of X is P ( X < x ) = 1 − e − x 9 P ( X < x ) = 1 − e − x 9 . Thus, P ( X > 20) = 1 - P ( X ≤ 20) = 1 − ( 1 − e − 20 9 ) ≈ 0.1084. 1 − ( 1 − e − 20 9 ) ≈ 0.1084.

We could also deduce that each person arriving has a 8/9 chance of not having Type B blood. So the probability that none of the first 20 people arrive have Type B blood is ( 8 9 ) 20 ≈ 0.0948 ( 8 9 ) 20 ≈ 0.0948 . (The geometric distribution is more appropriate than the exponential because the number of people between Type B people is discrete instead of continuous.)

Let T = duration (in minutes) between successive visits. Since patients arrive at a rate of one patient every seven minutes, μ = 7 and the decay constant is m = 1 7 1 7 . The cdf is P ( T < t ) = 1 − e t 7 1 − e t 7

• P ( T < 2) = 1 - 1 − e − 2 7 1 − e − 2 7 ≈ 0.2485.
• P ( T > 15) = 1 − P ( T < 15 ) = 1 − ( 1 − e − 15 7 ) ≈ e − 15 7 ≈ 0.1173 1 − P ( T < 15 ) = 1 − ( 1 − e − 15 7 ) ≈ e − 15 7 ≈ 0.1173 .
• P ( T > 15| T > 10) = P ( T > 5) = 1 − ( 1 − e − 5 7 ) = e − 5 7 ≈ 0.4895 1 − ( 1 − e − 5 7 ) = e − 5 7 ≈ 0.4895 .
• Let X = # of patients arriving during a half-hour period. Then X has the Poisson distribution with a mean of 30 7 30 7 , X ∼ Poisson ( 30 7 ) ( 30 7 ) . Find P ( X > 8) = 1 – P ( X ≤ 8) ≈ 0.0311.

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Unit 8: Probability & Statistics Name: _ Date:__Per:_Homework 3: Theoretical vs. Experimental Probability ** This is a 2-page document! ** Give each probability as a simplified fraction, decimal, and percent 1. A number between 1 and 3 is a) Find and compare the theoretical probability and chosen 30 times. Results are shown experimental probability of choosing a 2. in the table below. Theoretical: Experimental: Compare: 2. The spinner below is spun 60 times. a) Find and compare the theoretical probability and Results are shown in the table experimental probability of spinning an A

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Data and Statistics Unit 7th Grade CCSS

A 9 day CCSS-Aligned Statistics Unit – including populations and samples, drawing inferences from samples, measures of centers and variability, comparing and analyzing dot and box plots.

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