Question

8 Q099 Approximating a Binomial Distribution die 2 OOOO00 Probability Experiment: 2 die There are are 36 possible outcomes. Let X be a random variable. X = sum of the outcomes of the 2 die 0000 we roll the outcome times. You die 1 2 3 4 5 6 7 3 4 5 6 7 8 4 5 6 7 8 9 5 6 7 6 7 8 7 8 9 8 9 10) 9,10,11 10 11 12 probability The table to the left shows the 36 outcomes for X. If we roll the pair of die 220 times, what is the proba- bility that the outcome 10 occurs at most times 21 means 21 or less than 21 times. You must show the standard normal graph show the correct probability area). Round your final answer to 4 significant digits. Remember to use continuity correction. 10 points (graph is 4 of the 10 points) Hint : Calculate P(10). ::

Answer 1

Out of 36 outcomes, 3 outcomes show sum 10 so the probability of getting a sum 10 in a single roll of 2 die is

p = 3 / 36 = 1 / 12

Here X has binomial distribution with parameters as follow:

n = 220, p = 1/12

Using normal approximation, X has approximately normal distribution with mean and SD as follows:

$\mu=np=220\cdot \frac{1}{12}=18.333$

$\sigma=\sqrt{np(1-p)}=\sqrt{220\cdot \frac{1}{12}\cdot \frac{11}{12}}=4.0995$

Using continuity correction factor we need to find the probability as follow:

$P(X\leq 21)=P(X<21&plus;0.5)=P(X<21.5)$

The z-score for X = 21.5 is

$z=\frac{X-\mu}{\sigma}=\frac{21.5-18.333}{4.0995}=0.77$

The required probability is

$P(X\leq 21)=P(X<21&plus;0.5)=P(X<21.5)=P(z<0.77)=0.7794$

Following is the graph:

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