Question

Scrabble In the game of Scrabble, 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 an SRS of 7 tiles. Let $\hat{p}$be the proportion of vowels in her sample.

(a) Is the $10\%$condition met in this case? Justify your answer.

(b) Is the Normal condition met in this case? Justify your answer.

Short answer

a). Yes, is the $10\%$condition met in this case.

b). No, the Normal condition is met in this case.

Step-by-step solution

Part (a) Step 1: Given Information

Number of tiles $=100$

Vowels$=42$,

Consonants $=56$,

Number of blank tiles $=2$

Number of tiles for SRS$=7$

Part (a) Step 2: Explanation

Given a population of $100$people, a sample size of $7$represents less than ten percent of the total. This indicates that the ten percent criteria has been met. As a result, the condition is satisfied.

Part (b) Step 1: Given Information 

Number of tiles$=100$

Vowels$=42$

Consonants$=56$

Number of blank tiles$=2$

Number of tiles for SRS$=7$

Part (b) Step 2: Explanation 

Given for the population:

$x=42$

$n=100$

The population proportion is calculated by dividing the sample size by the number of successes:

localid="1650011266324" $$\begin{gathered} p=\frac{x}{n} \\ =\frac{42}{100} \end{gathered}$$

Sample given: $n=7$

As long as $np$and $n(1-p)$are both at least $10$, the condition is met.

localid="1650011270157" $$\begin{gathered} np=7\times \frac{42}{100} \\ \approx 2.94 \end{gathered}$$

localid="1650011273619" $$\begin{gathered} n(1-p)=7\times \left(1-\frac{42}{100}\right) \\ \approx 4.06 \end{gathered}$$

Neither of them exceeds $10$, so the Normal condition has not been met.