Math Problem Statement

According to the article 'Are Babies Normal?' by Traci Clemons and Marcello Pagano, the birth weights of babies are normally distributed with a mean of 3346 grams and a standard deviation of 560 grams. 1. What is the probability that a randomly selected baby weighs between 3100 grams and 3700 grams? Round your answer to 4 decimal places. 2. What is the probability that the average weight of 13 randomly selected babies is between 3100 grams and 3700 grams? Round your answer to 4 decimal places. 3. Why did the probability increase?

Solution

Let's go through each question step-by-step.

Given Information:

  • Mean (μ\mu) = 3346 grams
  • Standard deviation (σ\sigma) = 560 grams
  • The distribution is normally distributed.

Question 1:

What is the probability that a randomly selected baby weighs between 3100 grams and 3700 grams? Round your answer to 4 decimal places.

For a normal distribution, we can use the Z-score formula: Z=XμσZ = \frac{X - \mu}{\sigma} where:

  • XX is the value we’re standardizing,
  • μ\mu is the mean,
  • σ\sigma is the standard deviation.

Step 1: Find the Z-scores

  1. For 3100 grams: Z3100=31003346560=2465600.4393Z_{3100} = \frac{3100 - 3346}{560} = \frac{-246}{560} \approx -0.4393

  2. For 3700 grams: Z3700=37003346560=3545600.6321Z_{3700} = \frac{3700 - 3346}{560} = \frac{354}{560} \approx 0.6321

Step 2: Look up the Z-scores in a standard normal distribution table or use a calculator:

  • For Z=0.4393Z = -0.4393, the probability is approximately 0.3300.
  • For Z=0.6321Z = 0.6321, the probability is approximately 0.7367.

Step 3: Calculate the Probability

The probability that a randomly selected baby weighs between 3100 grams and 3700 grams is: P(3100<X<3700)=P(Z<0.6321)P(Z<0.4393)=0.73670.3300=0.4067P(3100 < X < 3700) = P(Z < 0.6321) - P(Z < -0.4393) = 0.7367 - 0.3300 = 0.4067

So, the answer to Question 1 is approximately 0.4067.


Question 2:

What is the probability that the average weight of 13 randomly selected babies is between 3100 grams and 3700 grams? Round your answer to 4 decimal places.

Since we’re dealing with the average of a sample, we need to adjust the standard deviation by dividing it by the square root of the sample size nn: σmean=σn=56013155.30\sigma_{\text{mean}} = \frac{\sigma}{\sqrt{n}} = \frac{560}{\sqrt{13}} \approx 155.30

Step 1: Find the Z-scores for the sample mean distribution

  1. For 3100 grams: Z3100=31003346155.30=246155.301.583Z_{3100} = \frac{3100 - 3346}{155.30} = \frac{-246}{155.30} \approx -1.583

  2. For 3700 grams: Z3700=37003346155.30=354155.302.279Z_{3700} = \frac{3700 - 3346}{155.30} = \frac{354}{155.30} \approx 2.279

Step 2: Look up the Z-scores

  • For Z=1.583Z = -1.583, the probability is approximately 0.0569.
  • For Z=2.279Z = 2.279, the probability is approximately 0.9887.

Step 3: Calculate the Probability

The probability that the average weight of 13 randomly selected babies is between 3100 grams and 3700 grams is: P(3100<Xˉ<3700)=P(Z<2.279)P(Z<1.583)=0.98870.0569=0.9318P(3100 < \bar{X} < 3700) = P(Z < 2.279) - P(Z < -1.583) = 0.9887 - 0.0569 = 0.9318

So, the answer to Question 2 is approximately 0.9318.


Question 3:

Why did the probability increase?

The probability increased because the sample size increased, which reduces the standard deviation of the sampling distribution. This makes the distribution of sample means more concentrated near the mean of 3346, increasing the likelihood that the average of 13 babies falls within the specified range.

Correct answer:

  • "The probability increased since the sample size increased and the sample means are more concentrated near the mean of 3346."

Would you like more details on any step, or have any questions?


Additional Questions to Consider:

  1. How would the probability change if the standard deviation were different?
  2. What would the Z-score be for a weight of 3000 grams?
  3. How would you calculate the probability for a weight above 3700 grams?
  4. What impact does increasing the sample size have on the sampling distribution?
  5. How can you interpret the probabilities in a real-world context?

Tip:

Always remember that increasing the sample size makes the sample mean more stable, reducing the variability (standard deviation) of the sampling distribution.

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Probability
Z-score
Sampling Distribution

Formulas

Z = (X - μ) / σ
σ_mean = σ / √n
P(a < X < b) = P(Z < Z_b) - P(Z < Z_a)

Theorems

Central Limit Theorem

Suitable Grade Level

College Level - Statistics