Question

A poll agency reports that 37% of teenagers aged 12-17 own smartphones. A random sample of 101 teenagers is drawn. Round your answers to at least four decimal places as needed.
A) find the mean u^p

Answers

Answer
find the standard deviation o^p
Answer
(c) Find the probability that more than 40% of the sampled teenagers own a smartphone.
Answer
(d) Find the probability that the proportion of the sampled teenagers who own a smartphone is between .34 and .48.
Answer
(e) Find the probability that less than 48% of sampled teenagers own smartphones.
Answer

(f) Would it be unusual if less than 30% of the sampled teenagers owned smartphones?
it [would or wouldnt?] since the probability is [?]
Answered by GPT-4o mini
To find the mean \( \mu_{p} \) of the sampling distribution of the sample proportion \( p \), you can use the formula:

\[
\mu_{p} = p
\]

where \( p \) is the population proportion.

In this case, you are given that 37% of teenagers aged 12-17 own smartphones, which can be expressed as:

\[
p = 0.37
\]

Thus, the mean \( \mu_{p} \) is:

\[
\mu_{p} = 0.37
\]

Therefore, the answer is:

\[
\mu_{p} = 0.3700
\] (rounded to four decimal places).
Answered by GPT-4o mini
To find the standard deviation \( \sigma_{p} \) of the sampling distribution of the sample proportion \( p \), you can use the formula:

\[
\sigma_{p} = \sqrt{\frac{p(1 - p)}{n}}
\]

where:
- \( p \) is the population proportion,
- \( n \) is the sample size.

Given:
- \( p = 0.37 \)
- \( n = 101 \)

We can calculate \( \sigma_{p} \) as follows:

1. Calculate \( 1 - p \):
\[
1 - p = 1 - 0.37 = 0.63
\]

2. Now plug in the values into the formula:
\[
\sigma_{p} = \sqrt{\frac{0.37 \times 0.63}{101}}
\]

3. Calculate the numerator:
\[
0.37 \times 0.63 = 0.2331
\]

4. Divide by the sample size \( n = 101 \):
\[
\frac{0.2331}{101} \approx 0.002308
\]

5. Take the square root:
\[
\sigma_{p} \approx \sqrt{0.002308} \approx 0.0480
\]

Thus, the standard deviation \( \sigma_{p} \) is approximately:

\[
\sigma_{p} \approx 0.0480
\] (rounded to four decimal places).
Answered by GPT-4o mini
To find the probability that more than 40% of the sampled teenagers own a smartphone, we can use the normal approximation to the binomial distribution. We will convert the sample proportion to a z-score and then use the standard normal distribution to find the probability.

### Step 1: Identify Parameters
- Population proportion (\( p \)) = 0.37
- Sample size (\( n \)) = 101
- Sample proportion threshold = 0.40

### Step 2: Calculate the Mean and Standard Deviation of the Sampling Distribution
We already calculated these in previous steps:
- Mean \( \mu_{p} = 0.37 \)
- Standard deviation \( \sigma_{p} \approx 0.0480 \)

### Step 3: Calculate the Z-Score
To calculate the z-score for the proportion of 0.40:

\[
z = \frac{p' - \mu_{p}}{\sigma_{p}}
\]

Where:
- \( p' = 0.40 \)
- \( \mu_{p} = 0.37 \)
- \( \sigma_{p} \approx 0.0480 \)

Substituting the values:

\[
z = \frac{0.40 - 0.37}{0.0480} \approx \frac{0.03}{0.0480} \approx 0.625
\]

### Step 4: Find the Probability
Next, we need to find the probability that corresponds to a z-score of 0.625. You can use a standard normal distribution table or a calculator.

Using a z-table or calculator, you can find that:

\[
P(Z < 0.625) \approx 0.7357
\]

However, since we want the probability that more than 40% own a smartphone, we need:

\[
P(Z > 0.625) = 1 - P(Z < 0.625) = 1 - 0.7357 \approx 0.2643
\]

### Final Answer
Thus, the probability that more than 40% of the sampled teenagers own a smartphone is:

\[
P(p' > 0.40) \approx 0.2643
\]
(rounded to four decimal places).
Answered by GPT-4o mini
To find the probability that the proportion of the sampled teenagers who own a smartphone is between 0.34 and 0.48, we will again use the normal approximation to the binomial distribution and calculate the z-scores for both proportions.

### Step 1: Identify Parameters
- Population proportion (\( p \)) = 0.37
- Sample size (\( n \)) = 101
- Proportions to analyze: \( p_1 = 0.34 \) and \( p_2 = 0.48 \)

### Step 2: Calculate the Mean and Standard Deviation of the Sampling Distribution
As calculated previously:
- Mean \( \mu_{p} = 0.37 \)
- Standard deviation \( \sigma_{p} \approx 0.0480 \)

### Step 3: Calculate the Z-Scores

**For \( p_1 = 0.34 \)**:
\[
z_1 = \frac{p_1 - \mu_p}{\sigma_p} = \frac{0.34 - 0.37}{0.0480} = \frac{-0.03}{0.0480} \approx -0.625
\]

**For \( p_2 = 0.48 \)**:
\[
z_2 = \frac{p_2 - \mu_p}{\sigma_p} = \frac{0.48 - 0.37}{0.0480} = \frac{0.11}{0.0480} \approx 2.2917
\]

### Step 4: Find the Corresponding Probabilities

**For \( z_1 = -0.625 \)**:
Using the standard normal distribution table or calculator, we find:
\[
P(Z < -0.625) \approx 0.2659
\]

**For \( z_2 = 2.2917 \)**:
Using the standard normal distribution table or calculator, we find:
\[
P(Z < 2.2917) \approx 0.9890
\]

### Step 5: Calculate the Probability Between the Two Z-Scores
To find the probability that the proportion of the sampled teenagers who own a smartphone is between 0.34 and 0.48:

\[
P(0.34 < p < 0.48) = P(Z < 2.2917) - P(Z < -0.625)
\]
\[
P(0.34 < p < 0.48) \approx 0.9890 - 0.2659 \approx 0.7231
\]

### Final Answer
Thus, the probability that the proportion of sampled teenagers who own a smartphone is between 0.34 and 0.48 is:

\[
P(0.34 < p < 0.48) \approx 0.7231
\]
(rounded to four decimal places).
Answered by GPT-4o mini
To find the probability that less than 48% of sampled teenagers own smartphones, we will again use the normal approximation to the binomial distribution and calculate the z-score for the proportion of 0.48.

### Step 1: Identify Parameters
- Population proportion (\( p \)) = 0.37
- Sample size (\( n \)) = 101
- Proportion to analyze: \( p = 0.48 \)

### Step 2: Calculate the Mean and Standard Deviation of the Sampling Distribution
We already have:
- Mean \( \mu_{p} = 0.37 \)
- Standard deviation \( \sigma_{p} \approx 0.0480 \)

### Step 3: Calculate the Z-Score

For \( p = 0.48 \):
\[
z = \frac{p - \mu_{p}}{\sigma_{p}} = \frac{0.48 - 0.37}{0.0480} = \frac{0.11}{0.0480} \approx 2.2917
\]

### Step 4: Find the Corresponding Probability

Using the standard normal distribution table or a calculator:
\[
P(Z < 2.2917) \approx 0.9890
\]

### Final Answer
Thus, the probability that less than 48% of sampled teenagers own smartphones is:

\[
P(p < 0.48) \approx 0.9890
\]
(rounded to four decimal places).
Answered by GPT-4o mini
To determine whether it would be unusual if less than 30% of the sampled teenagers owned smartphones, we need to calculate the probability of that event happening.

### Step 1: Calculate the Z-Score for \( p = 0.30 \)

We'll use the same mean \( \mu_{p} \) and standard deviation \( \sigma_{p} \) calculated earlier:
- Mean \( \mu_{p} = 0.37 \)
- Standard deviation \( \sigma_{p} \approx 0.0480 \)

For \( p = 0.30 \):
\[
z = \frac{p - \mu_{p}}{\sigma_{p}} = \frac{0.30 - 0.37}{0.0480} = \frac{-0.07}{0.0480} \approx -1.4583
\]

### Step 2: Find the Corresponding Probability

Using the standard normal distribution table or a calculator, we find:
\[
P(Z < -1.4583) \approx 0.0726
\]

### Conclusion

Since the probability \( P(p < 0.30) \approx 0.0726 \) is less than 0.05 (or 5%), this would be considered an unusual event.

Thus, we can conclude:

It **would be** unusual if less than 30% of the sampled teenagers owned smartphones since the probability is approximately **0.0726**.

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