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Brrunno [24]
3 years ago
6

Each card in a set of cards has one of the letters from the word MATHEMATICS. The cards are shuffled. Suppose you draw a card fr

om the set of cards, record the letter, return the card to the set, and shuffle the cards. You repeat this experiment 22 times. Would you expect the experimental probability of drawing a vowel to be the same as the theoretical probability? YES or NO and Explain you answer
Mathematics
1 answer:
Sergio039 [100]3 years ago
3 0

Answer:

No

Step-by-step explanation:

Theoretical probability is the expected probability of anything. For example we expect a probability of 1/6 for each number when we roll a die .

Experimental Probability is the one that is obtained by the actual experimenting of rolling the die.

The theoretical probability is the example of perfect probability and experimental probability  is the example of actual occurrence.

Their formulas are

Theoretical Probability = number of possible outcomes/ total number of possible outcomes

Experimental Probability =  number of favorable outcomes/ total number of actual outcomes

If we a roll a fair die several times we may get 2 repeated number of times or any other number that is on the die.

So the experimental probability is bit different from the theoretical probability.

Now if the cards are shuffled and the experiment is repeated 22 times

experimental probability of drawing a vowel  would not  be the same as the theoretical probability because n is small.

If n→1000 or more

n→∞

According to K. Pearson or Buffon when n tends to infinity the  theoretical probability becomes equal to experimental probability.

They experimented this on a coin and put forward their results.

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In Major League Baseball, the American League (AL) allows a designated hitter (DH) to bat in place of the pitcher, but in the Na
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Answer:

Step-by-step explanation:

From the given information:

Let:

\mu_1 represent the population mean no. for DH group.

\mu_2 represent the population mean no. for no DH group  

n_1 represent the sample sizes of DH

n_2 represent the sample sizes of  no DH

Sample size: n_1 = n_2 = 20

For both groups, the population standard deviation of runs scored = 2.54  

i.e

\sigma^2_1 = \sigma_2^2 = 2.54

The null & alternative hypothesis;

H_o : \mu_1 \le \mu_2 \\ \\ H_a: \mu_1 > \mu_2

Level of significance:

The test statistics is:

z = \dfrac{\bar x_1 - \bar x_2}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma_2^2}{n_2} }}

where;

\bar x_1 = sample mean for DH group

\implies \dfrac{1}{n_1} \sum \limit ^{n_1}_{i=1} x_1  \\ \\ = \dfrac{1}{20}(0+6+8+2+2+4+7+7+6+5+1+1+5+4+4+5+7+11+10+0) \\ \\ = \dfrac{92}{20}

= 4.6

\bar x_2 = sample mean for no DH group

\implies \dfrac{1}{n_2} \sum \limit ^{n_1}_{i=1} x_1  \\ \\ = \dfrac{1}{20}(3+6+2+4+0+5+7+6+1+8+12+4+6+3+4+0+5+2+1+4) \\ \\ = \dfrac{83}{20}

= 4.15

Now:

z = \dfrac{4.60- 4.15}{\sqrt{\dfrac{2.54^2}{20} + \dfrac{2.54^2}{20} }} \\ \\  = \dfrac{0.45}{\sqrt{0.32258 +0.32258 }} \\ \\ = \dfrac{0.45}{0.803219} \\ \\

= 0.5602

Since the test is one-tailed by looking at that H_a:

The P-value = P(Z > z)

\implies 1 - P(Z \le z) \\ \\ = 1 - P(Z \le 0.5602)

Using Excel Function " =normdist(z)"; we have":

= 1- 0.7123

P-value = 0.2877

Decision rule: To reject H_o, if p-value < \alpha  \ at \  0.10

Conclusion: SInce P = 0.2877 which is >  \ \alpha  \ at \  0.10.

We fail to reject the H_o and conclude that there is insufficient evidence to support the given claim that: \text{more runs are scored in games for which DH is used.}

 

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