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Morgarella [4.7K]
3 years ago
5

Erika is following a 2200 calorie per day diet.she eats the recommended servings of bread s and cereals, averaging 120 calories

per serving. She also eats 5 servings of vegetables. If the rest of her daily intake is 870 calories, what is the average numberof calories in eachserving of vegetables?
Mathematics
1 answer:
lora16 [44]3 years ago
7 0
The answer should be 50 calories in each serving of vegetables.

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Calculate the expected value, the variance, and the standard deviation of the given random variable X. (Round all answers to two
Tom [10]

Answer:

E (X) = 1.8

Var (X) = 0.36

σ = 0.6

Step-by-step explanation:

Solution:-

- Denote the random variable X : is the number of red marbles that Suzan has in her hand after she selects three marbles.

- Total sample space (bag) have the following quantity of colored marbles:

                Bag : { 3 Red , 2 Green }

- Suzan selects three marbles from the bag. The Event (X) defines the number of red marbles out of 3.

- The total number of outcomes / selections for randomly selecting 3 balls from the bag:

               All outcomes = 5 C 3 = 10

- The probability distribution of the random variable X, we will use combinations to determine the required probabilities:

  X = 1 red marble:

        P ( X = 1 ) : Suzan chooses 1 Red marble from the available 3 red marble and 2 green marbles.

        P ( X = 1 ) = [ 3C1*2C2 ] / all outcomes = (3*1) / 10 = 0.3

X = 2 red marble:

        P ( X = 2 ) : Suzan chooses 2 Red marble from the available 3 red marble and 1 green marbles.

        P ( X = 2 ) = [ 3C2*2C1 ] / all outcomes = (3*2) / 10 = 0.6

X = 3 red marble:

        P ( X = 3 ) : Suzan chooses 3 Red marble from the available 3 red marble.

        P ( X = 3 ) = [ 3C3] / all outcomes = (1) / 10 = 0.1

- The probability distribution is as follows:

          X :         1                 2                3

       P (X):       0.3            0.6              0.1

- The expected value E(X) for the given random variable X is:

                E ( X ) = ∑Xi*P(Xi)

                           = 1*0.3 + 2*0.6 + 3*0.1

                           =1.8

- The variance Var(X) for the given random variable X is:

                Var ( X ) = ∑Xi^2*P(Xi) - [ E(X) ] ^2

                               = 1^2*0.3 + 2^2*0.6 + 3^2*0.1 - 1.8^2

                               = 0.36

- The standard deviation for the given random variable X is:

                σ = √Var(X)

                σ = √0.36

                σ = 0.6

5 0
3 years ago
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