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In the game of Scrabble [tex]\(^{\circledR}\)[/tex], each player begins by drawing 7 tiles from a bag containing 100 tiles. There are 42 vowels, 56 consonants, and 2 blank tiles in the bag. Cait chooses a simple random sample (SRS) of 7 tiles. Let [tex]\(\hat{p}\)[/tex] be the proportion of vowels in her sample.

Is the Large Counts condition met in this case? Justify your answer.

A. The Large Counts condition is met because [tex]\(n p\)[/tex] and [tex]\(n(1-p)=51\)[/tex] are both at least 10.

B. The Large Counts condition is not met because the sample size was only [tex]\(n=7\)[/tex]. Neither [tex]\(n p\)[/tex] nor [tex]\(n(1-p)\)[/tex] will be at least 10.

C. The Large Counts condition is not met because the sample size is less than 30.

D. The Large Counts condition is met because [tex]\(n p=42\)[/tex] and [tex]\(n(1-p)=58\)[/tex] are both at least 10.

E. The Large Counts condition is not met because [tex]\(n p=7\)[/tex] and [tex]\(n(1-p)=35\)[/tex] are not both at least 10.

Answer :

To determine if the Large Counts condition is met for Cait’s sample of 7 Scrabble tiles, we need to consider the expected number of vowels and consonants (including blank tiles) in her sample. The Large Counts condition is typically used in statistical inference to ensure that sample sizes are large enough to use normal approximation methods.

Step 1: Calculate the proportion of vowels in the entire bag.
- Total number of tiles in the bag: 100
- Number of vowels: 42

The probability of drawing a vowel ([tex]\(p\)[/tex]) is the number of vowels divided by the total number of tiles:
[tex]\[ p = \frac{42}{100} = 0.42 \][/tex]

Step 2: Determine the expected counts for vowels and non-vowels in the sample:
- Sample size ([tex]\(n\)[/tex]): 7

Calculate the expected number of vowels ([tex]\(n \times p\)[/tex]):
[tex]\[ n \times p = 7 \times 0.42 = 2.94 \][/tex]

Calculate the expected number of non-vowels ([tex]\(n \times (1-p)\)[/tex]):
- Since there are 100 tiles in total, the probability of drawing a non-vowel is:
[tex]\[ 1 - p = 1 - 0.42 = 0.58 \][/tex]
[tex]\[ n \times (1-p) = 7 \times 0.58 = 4.06 \][/tex]

Step 3: Check the Large Counts condition:
- For the Large Counts condition to be met, both [tex]\(n \times p\)[/tex] (expected number of vowels) and [tex]\(n \times (1-p)\)[/tex] (expected number of non-vowels) should be at least 10.

Given:
[tex]\[ n \times p = 2.94 \][/tex]
[tex]\[ n \times (1-p) = 4.06 \][/tex]

Both expected counts, 2.94 and 4.06, are less than 10.

Conclusion:
The Large Counts condition is not met because neither the expected number of vowels (2.94) nor the expected number of non-vowels (4.06) in Cait's sample of 7 tiles reaches the minimum of 10 required for the condition to be satisfied.

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