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WARM – UP 1.) In a box of 10 batteries, 3 batteries are Dead. You choose 2 batteries from the box at random: a) Create the probability model for the.

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Presentation on theme: "WARM – UP 1.) In a box of 10 batteries, 3 batteries are Dead. You choose 2 batteries from the box at random: a) Create the probability model for the."— Presentation transcript:

1 WARM – UP 1.) In a box of 10 batteries, 3 batteries are Dead. You choose 2 batteries from the box at random: a) Create the probability model for the Number of good batteries you get. What is the expected number of good batteries you get on average. What is the standard deviation. 2.) The amount of Milk in a half gallon carton is a random variable with a mean of 1800 g and a standard deviation of 30 g. The amount of Milk that I drink a week mean of 520 g and a standard deviation of 15 g. If the weight of the Milk remaining after a week can be described by a normal model, what's the probability that the carton of Milk has more than 900 g after a week?

2 WARM – UP 1.) In a box of 10 batteries, 3 batteries are Dead. You choose 2 batteries from the box at random: a) Create the probability model for the Number of good batteries you get. What is the expected number of good batteries you get on average. What is the standard deviation. X P(X)

3 WARM – UP 2.) The amount of Milk in a half gallon carton is a random variable with a mean of 1800 g and a standard deviation of 30 g. The amount of Milk that I drink a week has a mean of 520 g and a standard deviation of 15 g. If the weight of the Milk remaining after a week can be described by a normal model, what's the probability that the carton of Milk has more than 1500 g after a week? P(xd > 1500) = 1800 – 520 μ(X – Y) = 1280 g σ(X – Y) = g = Normalcdf(6.559`, E99) = 0

4 WARM – UP #1. A company produces ceramic floor tiles which are supposed to have a surface area of 16 square inches. Due to variability in the manufacturing process, the actual surface area of all tiles has a normal distribution with mean 16.1 square inches and standard deviation 0.2 square inches. What is the proportion of tiles produced by the process with surface area less than 16.0 square inches? #2. The weight of a randomly selected can of a new soft drink is known to have a normal distribution with mean 8.3 ounces and standard deviation 0.2 ounces. What weight (in ounces) should be stamped on the can so that only 2% of the cans are underweight? P(x < 16) = ? P(z < - 0.5) = Normalcdf(-E99, -0.5) = P(x < ?) = .02 z = InvNorm(.02) = -2.05 ? = 7.89

5 NEW guideline specify that the INSPECTION time must be doubled.
The following are the times needed to completing assemble a motorcycle for road worthiness under old specifications. Phase Mean (min.) Std. D Assemble 85.2 6.7 Tune 26.5 4.4 Inspect 8.1 2.3 NEW guideline specify that the INSPECTION time must be doubled. a) What is the mean and std. deviation for the set up time? b) What is the probability that a motorcycle can be set up in less than two hours? µ = A + T + 2∙I σ2 = A 2 + T ∙I 2 P(Set up < 120) P(Set up < 120) =

6 RANDOM VARIABLES A Random Variable is a variable whose value is a numerical outcome of a random phenomenon. Usually denoted by X. A Discrete Random Variable, X, has a finite or countable number of possible values. The Probability Distribution of X lists the values and their respective probabilities. EXAMPLE: The following probability distribution represent the AP Statistics scores from previous years. AP Score 1 2 3 4 5 Probability 0.14 0.24 0.34 0.21 0.07 1. All probabilities are: 0 ≤ p(x) ≤ 1 2. The sum of all probabilities equals one: Σ pi(x) = 1

7 A Continuous Random Variable, X, takes on all the infinite numerical values within an interval. The Probability Distribution of X is described by a Density curve. The probability of any event is the area under the density curve. EXAMPLE #1: UNIFORM Density Curve Let X represent any random number between 0 and 5. Find the following Probabilities: ? A= B·h 1 = 5·h 1/5 = 0.2 1. What is the Probability of any one number occurring? P( 0 ≤ X ≤ 1.2) = P( 0.4 ≤ X ≤ 3.5) = P(X > 2.2) = P(X ≥ 2.2) = 0.24 0.62 0.56

8 EXAMPLE #2: NORMAL Density Curve
Let X represent any random number from the normal distribution with mean 2.5 and standard deviation 0.8. 1. P( 0 ≤ X ≤ 1.2) = P( 0.4 ≤ X ≤ 3.5) = P(X > 2.2) = P(X ≥ 2.2) = 0.0512 0.8900 Normalcdf (zL, zU) 0.6462 OR Normalcdf (xL, xU, μ, σ)


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