Education professionals refer to science, technology, engineering, and mathematics as the STEM disciplines. A research group reported that 27% of freshmen entering college in a recent year planned to major in a STEM discipline. A random sample of 80 freshmen is selected. Round the answer to at least four decimal places.

Question

Is it appropriate to use the normal approximation to find the probability that less than 28% of the freshmen in the sample are planning to major in a STEM discipline?
It [is or is not?] appropriate to use the normal curvce since np=? [< or >?] 10 and n(1-p)=? {< or >] 10

Answer 1

To determine whether it is appropriate to use the normal approximation for the binomial distribution in this scenario, we need to check the conditions involving \(np\) and \(n(1-p)\).

1. Calculate \(p\): The proportion of freshmen planning to major in a STEM discipline is given as \(p = 0.27\).

2. Calculate \(n\): The sample size is \(n = 80\).

3. Calculate \(np\): \[ np = 80 \times 0.27 = 21.6 \]

4. Calculate \(n(1-p)\): \[ n(1-p) = 80 \times (1 - 0.27) = 80 \times 0.73 = 58.4 \]

Now we check the conditions:

• \(np = 21.6 > 10\)
• \(n(1-p) = 58.4 > 10\)

Since both \(np\) and \(n(1-p)\) are greater than 10, it is appropriate to use the normal approximation.

Therefore, it is appropriate to use the normal curve since \(np > 10\) and \(n(1-p) > 10\).

Answer 2

Find the probability that less than 28% of the freshmen in the sample are planning to major in a STEM discipline.

Answer 3

To find the probability that less than 28% of the freshmen in the sample are planning to major in a STEM discipline, we can use the normal approximation to the binomial distribution.

Given:

• Sample size \( n = 80 \)
• Proportion \( p = 0.27 \)
• We want to find the probability \( P(\hat{p} < 0.28) \), where \( \hat{p} \) is the sample proportion.

Step 1: Calculate the mean and standard deviation of the sample proportion \( \hat{p} \).

The mean \( \mu_{\hat{p}} \) of the sampling distribution of the sample proportion is given by: \[ \mu_{\hat{p}} = p = 0.27 \]

The standard deviation \( \sigma_{\hat{p}} \) is given by: \[ \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.27 \times 0.73}{80}} \approx \sqrt{\frac{0.1971}{80}} \approx \sqrt{0.00246375} \approx 0.0496 \]

Step 2: Convert the sample proportion to a z-score.

We want to find the z-score for \( \hat{p} = 0.28 \): \[ z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.28 - 0.27}{0.0496} \approx \frac{0.01}{0.0496} \approx 0.2016 \]

Step 3: Find the probability using the z-score.

Now we look up this z-score in the standard normal distribution table or use a calculator to find \( P(Z < 0.2016) \).

Using a z-table or calculator: \[ P(Z < 0.2016) \approx 0.5793 \]

Conclusion:

The probability that less than 28% of the freshmen in the sample are planning to major in a STEM discipline is approximately 0.5793 (or 57.93%).

Answer 4

A new sample of 130 freshmen is selected. Find the probability that less than 28% of the freshmen in this new sample are planning to major in a STEM discipline.

Answer 5

To find the probability that less than 28% of a new sample of 130 freshmen are planning to major in a STEM discipline, we will again use the normal approximation to the binomial distribution.

Given:

• Sample size (\(n\)) = 130
• Population proportion (\(p\)) = 0.27 (the proportion of all freshmen)

Step 1: Calculate the mean and standard deviation of the sample proportion (\(\hat{p}\)).

The mean \(\mu_{\hat{p}}\) of the sampling distribution of the sample proportion is given by: \[ \mu_{\hat{p}} = p = 0.27 \]

The standard deviation \(\sigma_{\hat{p}}\) is given by: \[ \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.27 \times 0.73}{130}} \approx \sqrt{\frac{0.1971}{130}} \approx \sqrt{0.001515} \approx 0.0389 \]

Step 2: Convert the sample proportion to a z-score.

We want to find the z-score for \(\hat{p} = 0.28\): \[ z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.28 - 0.27}{0.0389} \approx \frac{0.01}{0.0389} \approx 0.256 \]

Step 3: Find the probability using the z-score.

Now we look up this z-score in the standard normal distribution table or use a calculator to find \(P(Z < 0.256)\).

Using a z-table or calculator: \[ P(Z < 0.256) \approx 0.6016 \]

Conclusion:

The probability that less than 28% of the freshmen in the new sample of 130 are planning to major in a STEM discipline is approximately 0.6016 (or 60.16%).

Answer 6

Find the probability that the proportion of freshmen in the sample of 130 who plan to major in a STEM discipline is between .31 and .35.

Answer 7

To find the probability that the proportion of freshmen in a sample of 130 who plan to major in a STEM discipline is between 0.31 and 0.35, we can again use the normal approximation.

Given:

• Sample size (\(n\)) = 130
• Population proportion (\(p\)) = 0.27 (the proportion of all freshmen)

Step 1: Calculate the mean and standard deviation of the sample proportion (\(\hat{p}\)).

1. Mean of the sample proportion: \[ \mu_{\hat{p}} = p = 0.27 \]

2. Standard deviation of the sample proportion: \[ \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.27 \times 0.73}{130}} \approx \sqrt{\frac{0.1971}{130}} \approx \sqrt{0.001515} \approx 0.0389 \]

Step 2: Convert the sample proportions to z-scores.

For \( \hat{p} = 0.31 \): \[ z_1 = \frac{0.31 - 0.27}{0.0389} \approx \frac{0.04}{0.0389} \approx 1.029 \]

For \( \hat{p} = 0.35 \): \[ z_2 = \frac{0.35 - 0.27}{0.0389} \approx \frac{0.08}{0.0389} \approx 2.058 \]

Step 3: Find the probabilities using the z-scores.

Now look up \(P(Z < z_1)\) and \(P(Z < z_2)\):

1. For \(z_1 = 1.029\):

   • Using a standard normal table or calculator: \[ P(Z < 1.029) \approx 0.8485 \]

2. For \(z_2 = 2.058\):

   • Using a standard normal table or calculator: \[ P(Z < 2.058) \approx 0.9803 \]

Step 4: Calculate the required probability.

The probability that the proportion of freshmen in the sample who plan to major in a STEM discipline is between 0.31 and 0.35: \[ P(0.31 < \hat{p} < 0.35) = P(Z < 2.058) - P(Z < 1.029) \approx 0.9803 - 0.8485 = 0.1318 \]

Conclusion:

The probability that the proportion of freshmen in the sample of 130 who plan to major in a STEM discipline is between 0.31 and 0.35 is approximately 0.1318 (or 13.18%).

Answer 8

Find the probability that more than 32% of the freshmen in the sample of 130 are planning to major in a STEM discipline.

Answer 9

To find the probability that more than 32% of the freshmen in a sample of 130 are planning to major in a STEM discipline, we can use the normal approximation to the binomial distribution.

Given:

• Sample size (\(n\)) = 130
• Population proportion (\(p\)) = 0.27 (the proportion of all freshmen)

Step 1: Calculate the mean and standard deviation of the sample proportion (\(\hat{p}\)).

1. Mean of the sample proportion: \[ \mu_{\hat{p}} = p = 0.27 \]

2. Standard deviation of the sample proportion: \[ \sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.27 \times 0.73}{130}} \approx \sqrt{\frac{0.1971}{130}} \approx \sqrt{0.001515} \approx 0.0389 \]

Step 2: Convert the sample proportion to a z-score.

For \(\hat{p} = 0.32\): \[ z = \frac{0.32 - 0.27}{0.0389} \approx \frac{0.05}{0.0389} \approx 1.287 \]

Step 3: Find the probability using the z-score.

Now, let's find \(P(Z > 1.287)\).

Using a standard normal distribution table or a calculator:

1. First, find \(P(Z < 1.287)\): \[ P(Z < 1.287) \approx 0.8997 \]

2. Now, find the probability \(P(Z > 1.287)\): \[ P(Z > 1.287) = 1 - P(Z < 1.287) \approx 1 - 0.8997 \approx 0.1003 \]

Conclusion:

The probability that more than 32% of the freshmen in the sample of 130 are planning to major in a STEM discipline is approximately 0.1003 (or 10.03%).

Answer 10

Would it be unusual if less than 22% of the freshmen in the sample of 130 were planning to major in a STEM discipline?
It [would or would not] be unusual since the probability is [?]

Answer 11

To determine whether it would be unusual if less than 22% of the freshmen in the sample of 130 were planning to major in a STEM discipline, we can calculate the probability of this occurrence and then interpret that probability.

Given:

• Sample size (\(n\)) = 130
• Population proportion (\(p\)) = 0.27

Step 1: Calculate the mean and standard deviation of the sample proportion (\(\hat{p}\)).

1. Mean of the sample proportion: \[ \mu_{\hat{p}} = p = 0.27 \]

2. Standard deviation of the sample proportion: \[ \sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.27 \times 0.73}{130}} \approx \sqrt{\frac{0.1971}{130}} \approx \sqrt{0.001515} \approx 0.0389 \]

Step 2: Convert the sample proportion to a z-score.

For \(\hat{p} = 0.22\): \[ z = \frac{0.22 - 0.27}{0.0389} \approx \frac{-0.05}{0.0389} \approx -1.287 \]

Step 3: Find the probability using the z-score.

Now we will find \(P(Z < -1.287)\).

Using a standard normal distribution table or calculator:

1. First, find \(P(Z < -1.287)\): \[ P(Z < -1.287) \approx 0.1003 \]

Conclusion:

Since \(P(Z < -1.287) \approx 0.1003\), this means there is about a 10.03% chance that less than 22% of the freshmen are planning to major in a STEM discipline.

• In most statistical contexts, a probability of less than 5% (or, sometimes, less than 1%) is considered unusual. Since 10.03% is greater than 5%, we conclude that it would not be unusual for less than 22% of the freshmen in the sample of 130 to be planning to major in a STEM discipline.

Therefore, it would not be unusual since the probability is approximately 0.1003 (or 10.03%).