Chapter 7.2, Problem 32E

(a)

To determine

Whether the 10% condition is met or not.

Answer to Problem 32E

The 10% condition is met.

Explanation of Solution

Given information:

Number of tiles = 100

Number of vowels = 42

Number of consonants = 56

Number of blank tiles = 2

Number of tiles for SRS = 7

Given the population size is 100.

Here, the sample size of 7 is less than 10% of the population size 100.

Hence, the 10% condition is met.

(b)

To determine

Whether the Normal condition is met or not.

Answer to Problem 32E

The Normal condition has not been met.

Explanation of Solution

Given information:

Number of tiles = 100

Number of vowels = 42

Number of consonants = 56

Number of blank tiles = 2

Number of tiles for SRS = 7

Formula used:

When the product of sample size and the sampling proportion that is, $np$ and $n\left(1-p\right)$ both are less than at least 10 then the sampling distribution is approximately Normal.

Consider that the sample distribution be p and sample size for SRS be n.

  $n=100$ and $x=42$

Find the population proportion by dividing the number successes by the sample size that is, $p=\frac{x}{n}$

Substitute 42 for x and 100 for n in the above expression.

  $p=\frac{42}{100}$

Know that the product of sample size and the sampling proportion that is, $np$ and $n\left(1-p\right)$ both are less than at least 10 then the sampling distribution is approximately Normal.

Substitute $\frac{42}{100}$ for p and 7 for n in the expression $np$ .

  $7\times \frac{42}{100}\approx 2.94$

Again, substitute $\frac{42}{100}$ for p and 7 for n in the expression $n\left(1-p\right)$ .

  $7\left(1-\frac{42}{100}\right)\approx 4.06$

It is seen that both $np$ and $n\left(1-p\right)$ are less than 10; so the condition for Normal distribution has not been met.

Hence, the Normal condition has not been met.