Fifty draws are made at random with replacement from the box [ 0 0 1 1 1]. There are 33 ticket 1’s among the draws. The expected value for the sum is , the observed value is , the chance error is , and the standard error is . Fill in the blanks. Hint: Chance error = ObservedOutcome – ExpectedOutcome.

- The Expected value for the sum is 30.

- The Observed sum of 50 draws is 33.

- The Chance Error on the 50 draws is 3.

- The Standard Error on 50 draws is 2.191.

Step-by-step explanation:

The box contains [0, 0, 1, 1, 1]

Using probability to predict the expected outcome.

On one draw, the probability of drawing a 0 is (2/5).

And the probability of drawing a 1 is (3/5).

Probability mass function would look like

X | P(X)

0 | 0.40

1 | 0.60

So, expected value on one draw would be

E(X) = Σ xᵢpᵢ

xᵢ = each variable

pᵢ = probability of each variable

E(X) = (0×0.40) + (1×0.60) = 0.60.

Standard error on one draw = √[Σ(xᵢ - μ)²/N]

μ = E(X) = 0.60

Σ(xᵢ - μ)² = (0 - 0.60)² + (0 - 0.60)² + (1 - 0.6)² + (1 - 0.6)² + (1 - 0.6)² = 1.20

SE = √(1.2/5) = 0.490

So, for 50 draws (with replacement),

E(50X) = 50E(X) = 50 × 0.60 = 30.

For 50 draws, standard error = √50 × 0.490 = 2.191

The expected value for the sum = 30

The observed valued for the sum = (33×1) + (17×0) = 33

Chance Error = (Observed Outcome) - (Expected Outcome) = 33 - 30 = 3

Standard error gives an idea of how large the chance error would be.

Standard error on 50 draws = 2.191

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