Adn instritors at a la ge university went to estimate the percentage of studerts who are bte to class: In a random sample of 36 students, it wes found that 15 had bean late to class at least once in the bost week

Define the indon variatle chooseyour answer-

Cissify the random variatle
chooseyour answer_

GhedingCondtions
B the sarplerandom?
choose your ansmer_

Arebothmpandn( P at leastten? chooseyour answer.
EIt reasonatle to syy that the sarplei’sless than 10% of the popubtion? chooseyour answer.
Hhe the conditions beensitiffed? chooseyour answer.

Marginffirrar
Find the margin of error at the confidencelevel State the answer to four decinal places 35 in 0.1234
MOE thpeyour ansmer.

Ask by Santiago Bolton. in the United States

Mar 19,2025

Question

Adn instritors at a la ge university went to estimate the percentage of studerts who are bte to class: In a random sample of 36 students, it wes found that 15 had bean late to class at least once in the bost week

Define the indon variatle chooseyour answer-

Cissify the random variatle
chooseyour answer_

GhedingCondtions
B the sarplerandom?
choose your ansmer_

Arebothmpandn( P at leastten? chooseyour answer.
EIt reasonatle to syy that the sarplei’sless than 10% of the popubtion? chooseyour answer.
Hhe the conditions beensitiffed? chooseyour answer.

Marginffirrar
Find the margin of error at the confidencelevel State the answer to four decinal places 35 in 0.1234
MOE thpeyour ansmer.

Ask by Santiago Bolton. in the United States

Mar 19,2025

UpStudy AI · Accepted Answer

Let's break down the problem step by step.

### Step 1: Define the random variable $$ \mathrm{X} $$

The random variable $$ \mathrm{X} $$ represents the number of students in the sample who have been late to class at least once in the past week.

So, we can define:
$$
\mathrm{X} = \text{the number of students who have been late to class at least once in the past week}
$$

### Step 2: Classify the random variable

The random variable $$ \mathrm{X} $$ is a discrete random variable because it can take on a finite number of values (in this case, from 0 to 36).

### Step 3: Check the conditions

1. **Is the sample random?**
   - Yes, the problem states that it is a random sample of 36 students.

2. **Are both $$ np $$ and $$ n(1 - p) $$ at least 10?**
   - First, we need to calculate $$ p $$, the proportion of students who have been late. From the sample, we have:
   $$
   p = \frac{15}{36}
   $$
   Now, we can calculate $$ np $$ and $$ n(1 - p) $$:
   $$
   n = 36, \quad p = \frac{15}{36}, \quad 1 - p = 1 - \frac{15}{36} = \frac{21}{36}
   $$
   Now calculate:
   $$
   np = 36 \cdot \frac{15}{36} = 15
   $$
   $$
   n(1 - p) = 36 \cdot \frac{21}{36} = 21
   $$
   Both $$ np $$ and $$ n(1 - p) $$ are greater than 10.

3. **Is it reasonable to say that the sample is less than 10% of the population?**
   - Yes, unless the population of students at the university is very small, a sample of 36 students is likely to be less than 10% of the total student population.

4. **Have the conditions been satisfied?**
   - Yes, all conditions for using the normal approximation to the binomial distribution have been satisfied.

### Step 4: Find the margin of error at the $$ 90\% $$ confidence level

To find the margin of error (MOE), we can use the formula:
$$
\text{MOE} = z \cdot \sqrt{\frac{p(1 - p)}{n}}
$$
Where $$ z $$ is the z-score corresponding to the desired confidence level. For a $$ 90\% $$ confidence level, the z-score is approximately $$ 1.645 $$.

Now, we can calculate $$ p $$ and $$ 1 - p $$:
$$
p = \frac{15}{36} \approx 0.4167, \quad 1 - p \approx 0.5833
$$

Now, we can calculate the standard error (SE):
$$
\text{SE} = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.4167 \cdot 0.5833}{36}}
$$

Let's calculate the SE and then the MOE.

### Calculating the SE and MOE

1. Calculate the SE:
$$
\text{SE} = \sqrt{\frac{0.4167 \cdot 0.5833}{36}} \approx \sqrt{\frac{0.2431}{36}} \approx \sqrt{0.00675} \approx 0.0822
$$

2. Now calculate the MOE:
$$
\text{MOE} = 1.645 \cdot 0.0822 \approx 0.1355
$$

### Final Answer

The margin of error at the $$ 90\% $$ confidence level is:
$$
\text{MOE} \approx 0.1355
$$

Thus, the answers to your questions are:

- $$ \mathrm{X} = \text{the number of students who have been late to class at least once in the past week} $$
- Random variable classification: Discrete
- Sample random: Yes
- $$ np $$ and $$ n(1 - p) $$ at least 10: Yes
- Sample less than 10% of the population: Yes
- Conditions satisfied: Yes
- Margin of Error $$ \text{MOE} = 0.1355 $$ (to four decimal places: 0.1355)
- **Random Variable $$ \mathrm{X} $$:** The number of students who have been late to class at least once in the past week. - **Classification:** Discrete. - **Sample Random:** Yes. - **$$ np $$ and $$ n(1 - p) $$ at least 10:** Yes. - **Sample Less Than 10% of Population:** Yes. - **Conditions Satisfied:** Yes. - **Margin of Error at 90% Confidence Level:** 0.1355

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