Question
The weight of male babies less than 2 months old in the United States is normally distributed with mean 12.4 and standard deviation 5.2. ) What proportion of babies weigh more than 14 pounds?
Answers
Answer
(b) What proportion of babies weigh less than 15 pounds
Answer
(c) What proportion of babies weigh between 10.3 and 14.9 pounds ?
Answer
(d) Is it unusual for a baby to weigh more than 17 pounds ? why or why not
Answered by
GPT-4o mini
To determine the proportion of male babies less than 2 months old who weigh more than 14 pounds, we can use the properties of the normal distribution.
1. **Given Information:**
- Mean (\(\mu\)) = 12.4 pounds
- Standard Deviation (\(\sigma\)) = 5.2 pounds
- We want to find the proportion of babies with weights greater than 14 pounds.
2. **Calculate the Z-score:**
The Z-score can be calculated using the formula:
\[
Z = \frac{(X - \mu)}{\sigma}
\]
where:
- \(X\) is the value we are interested in (14 pounds)
- \(\mu\) is the mean (12.4)
- \(\sigma\) is the standard deviation (5.2)
Now let's calculate the Z-score:
\[
Z = \frac{(14 - 12.4)}{5.2} = \frac{1.6}{5.2} \approx 0.3077
\]
3. **Find the Proportion:**
To find the proportion of babies weighing more than 14 pounds, we need to look up the Z-score in the standard normal distribution table, or use a calculator or software that provides the cumulative distribution function (CDF) for the normal distribution.
The CDF will give us the area to the left of the Z-score.
- Looking up \(Z = 0.3077\), we find that the area to the left of this Z-score is approximately 0.6217 (this value may vary slightly depending on the table or software used).
4. **Calculate the Proportion Above 14 Pounds:**
Since we are interested in the proportion that is greater than 14 pounds, we take the complement of the CDF:
\[
P(X > 14) = 1 - P(X \leq 14)
\]
\[
P(X > 14) = 1 - 0.6217 = 0.3783
\]
**Conclusion:** Approximately 37.83% of male babies less than 2 months old weigh more than 14 pounds.
1. **Given Information:**
- Mean (\(\mu\)) = 12.4 pounds
- Standard Deviation (\(\sigma\)) = 5.2 pounds
- We want to find the proportion of babies with weights greater than 14 pounds.
2. **Calculate the Z-score:**
The Z-score can be calculated using the formula:
\[
Z = \frac{(X - \mu)}{\sigma}
\]
where:
- \(X\) is the value we are interested in (14 pounds)
- \(\mu\) is the mean (12.4)
- \(\sigma\) is the standard deviation (5.2)
Now let's calculate the Z-score:
\[
Z = \frac{(14 - 12.4)}{5.2} = \frac{1.6}{5.2} \approx 0.3077
\]
3. **Find the Proportion:**
To find the proportion of babies weighing more than 14 pounds, we need to look up the Z-score in the standard normal distribution table, or use a calculator or software that provides the cumulative distribution function (CDF) for the normal distribution.
The CDF will give us the area to the left of the Z-score.
- Looking up \(Z = 0.3077\), we find that the area to the left of this Z-score is approximately 0.6217 (this value may vary slightly depending on the table or software used).
4. **Calculate the Proportion Above 14 Pounds:**
Since we are interested in the proportion that is greater than 14 pounds, we take the complement of the CDF:
\[
P(X > 14) = 1 - P(X \leq 14)
\]
\[
P(X > 14) = 1 - 0.6217 = 0.3783
\]
**Conclusion:** Approximately 37.83% of male babies less than 2 months old weigh more than 14 pounds.
Answered by
GPT-4o mini
To find the proportion of male babies less than 2 months old who weigh less than 15 pounds, we can again use the properties of the normal distribution.
1. **Given Information:**
- Mean (\(\mu\)) = 12.4 pounds
- Standard Deviation (\(\sigma\)) = 5.2 pounds
- We're interested in the weight \(X = 15\) pounds.
2. **Calculate the Z-score:**
The Z-score is calculated using the formula:
\[
Z = \frac{(X - \mu)}{\sigma}
\]
Substituting the values:
\[
Z = \frac{(15 - 12.4)}{5.2} = \frac{2.6}{5.2} \approx 0.5
\]
3. **Find the Proportion:**
Next, we look up the Z-score in the standard normal distribution table or use a calculator that provides the cumulative distribution function (CDF) for the normal distribution.
- Finding the CDF for \(Z = 0.5\), we can determine that the area to the left (the proportion of babies weighing less than 15 pounds) is approximately 0.6915.
4. **Conclusion:**
Therefore, the proportion of male babies less than 2 months old that weigh less than 15 pounds is approximately **69.15%**.
1. **Given Information:**
- Mean (\(\mu\)) = 12.4 pounds
- Standard Deviation (\(\sigma\)) = 5.2 pounds
- We're interested in the weight \(X = 15\) pounds.
2. **Calculate the Z-score:**
The Z-score is calculated using the formula:
\[
Z = \frac{(X - \mu)}{\sigma}
\]
Substituting the values:
\[
Z = \frac{(15 - 12.4)}{5.2} = \frac{2.6}{5.2} \approx 0.5
\]
3. **Find the Proportion:**
Next, we look up the Z-score in the standard normal distribution table or use a calculator that provides the cumulative distribution function (CDF) for the normal distribution.
- Finding the CDF for \(Z = 0.5\), we can determine that the area to the left (the proportion of babies weighing less than 15 pounds) is approximately 0.6915.
4. **Conclusion:**
Therefore, the proportion of male babies less than 2 months old that weigh less than 15 pounds is approximately **69.15%**.
Answered by
GPT-4o mini
To find the proportion of male babies less than 2 months old who weigh between 10.3 and 14.9 pounds, we will calculate the Z-scores for both weight values and then find the corresponding probabilities using the standard normal distribution.
### Given Information:
- Mean (\(\mu\)) = 12.4 pounds
- Standard Deviation (\(\sigma\)) = 5.2 pounds
### Step 1: Calculate the Z-scores
1. **For 10.3 pounds:**
\[
Z_{10.3} = \frac{(10.3 - 12.4)}{5.2} = \frac{-2.1}{5.2} \approx -0.4038
\]
2. **For 14.9 pounds:**
\[
Z_{14.9} = \frac{(14.9 - 12.4)}{5.2} = \frac{2.5}{5.2} \approx 0.4808
\]
### Step 2: Find the Proportions
Next, we need to look up the Z-scores in the standard normal distribution table or use a calculator.
1. **Looking up \(Z_{10.3} \approx -0.4038\)**:
- The cumulative probability \(P(Z < -0.4038)\) is approximately **0.3424**.
2. **Looking up \(Z_{14.9} \approx 0.4808\)**:
- The cumulative probability \(P(Z < 0.4808)\) is approximately **0.6844**.
### Step 3: Calculate the Proportion Between the Two Values
To find the proportion of babies weighing between 10.3 and 14.9 pounds, we can subtract the two probabilities:
\[
P(10.3 < X < 14.9) = P(Z < 0.4808) - P(Z < -0.4038)
\]
\[
P(10.3 < X < 14.9) = 0.6844 - 0.3424 = 0.3420
\]
### Conclusion
Therefore, the proportion of male babies less than 2 months old who weigh between 10.3 and 14.9 pounds is approximately **34.20%**.
### Given Information:
- Mean (\(\mu\)) = 12.4 pounds
- Standard Deviation (\(\sigma\)) = 5.2 pounds
### Step 1: Calculate the Z-scores
1. **For 10.3 pounds:**
\[
Z_{10.3} = \frac{(10.3 - 12.4)}{5.2} = \frac{-2.1}{5.2} \approx -0.4038
\]
2. **For 14.9 pounds:**
\[
Z_{14.9} = \frac{(14.9 - 12.4)}{5.2} = \frac{2.5}{5.2} \approx 0.4808
\]
### Step 2: Find the Proportions
Next, we need to look up the Z-scores in the standard normal distribution table or use a calculator.
1. **Looking up \(Z_{10.3} \approx -0.4038\)**:
- The cumulative probability \(P(Z < -0.4038)\) is approximately **0.3424**.
2. **Looking up \(Z_{14.9} \approx 0.4808\)**:
- The cumulative probability \(P(Z < 0.4808)\) is approximately **0.6844**.
### Step 3: Calculate the Proportion Between the Two Values
To find the proportion of babies weighing between 10.3 and 14.9 pounds, we can subtract the two probabilities:
\[
P(10.3 < X < 14.9) = P(Z < 0.4808) - P(Z < -0.4038)
\]
\[
P(10.3 < X < 14.9) = 0.6844 - 0.3424 = 0.3420
\]
### Conclusion
Therefore, the proportion of male babies less than 2 months old who weigh between 10.3 and 14.9 pounds is approximately **34.20%**.
Answered by
GPT-4o mini
To determine whether it is unusual for a baby less than 2 months old to weigh more than 17 pounds, we can calculate the Z-score for 17 pounds and then find the corresponding cumulative probability. In general, if a measurement falls beyond 2 standard deviations from the mean, it is often considered unusual (or an outlier).
### Given Information:
- Mean (\(\mu\)) = 12.4 pounds
- Standard Deviation (\(\sigma\)) = 5.2 pounds
- Weight of interest (\(X = 17\) pounds)
### Step 1: Calculate the Z-score
Using the formula for Z-score:
\[
Z = \frac{(X - \mu)}{\sigma}
\]
Substituting the values:
\[
Z = \frac{(17 - 12.4)}{5.2} = \frac{4.6}{5.2} \approx 0.8846
\]
### Step 2: Find the Cumulative Probability
Next, we can look up the Z-score in the standard normal distribution table or use a calculator to find the cumulative probability.
- The cumulative probability for \(Z \approx 0.8846\) is approximately **0.8120**. This means that about 81.20% of babies weigh less than 17 pounds.
### Step 3: Find the Proportion of Babies Weighing More Than 17 Pounds
To find the proportion of babies that weigh more than 17 pounds, we can subtract this cumulative probability from 1:
\[
P(X > 17) = 1 - P(X \leq 17)
\]
\[
P(X > 17) = 1 - 0.8120 = 0.1880
\]
### Step 4: Interpret the Result
A proportion of 0.1880 (or 18.80%) means that around 18.80% of babies weigh more than 17 pounds.
### Conclusion
- **Is it unusual?**
- While it may not be common for a baby to weigh more than 17 pounds (as approximately 18.80% of babies do), it does not fall beyond the typical threshold of 2 standard deviations from the mean. A Z-score of approximately 0.8846 indicates that 17 pounds is actually within the range of typical weights for babies less than 2 months old. Thus, while it is on the higher side, it is not statistically unusual for a baby to weigh more than 17 pounds.
### Given Information:
- Mean (\(\mu\)) = 12.4 pounds
- Standard Deviation (\(\sigma\)) = 5.2 pounds
- Weight of interest (\(X = 17\) pounds)
### Step 1: Calculate the Z-score
Using the formula for Z-score:
\[
Z = \frac{(X - \mu)}{\sigma}
\]
Substituting the values:
\[
Z = \frac{(17 - 12.4)}{5.2} = \frac{4.6}{5.2} \approx 0.8846
\]
### Step 2: Find the Cumulative Probability
Next, we can look up the Z-score in the standard normal distribution table or use a calculator to find the cumulative probability.
- The cumulative probability for \(Z \approx 0.8846\) is approximately **0.8120**. This means that about 81.20% of babies weigh less than 17 pounds.
### Step 3: Find the Proportion of Babies Weighing More Than 17 Pounds
To find the proportion of babies that weigh more than 17 pounds, we can subtract this cumulative probability from 1:
\[
P(X > 17) = 1 - P(X \leq 17)
\]
\[
P(X > 17) = 1 - 0.8120 = 0.1880
\]
### Step 4: Interpret the Result
A proportion of 0.1880 (or 18.80%) means that around 18.80% of babies weigh more than 17 pounds.
### Conclusion
- **Is it unusual?**
- While it may not be common for a baby to weigh more than 17 pounds (as approximately 18.80% of babies do), it does not fall beyond the typical threshold of 2 standard deviations from the mean. A Z-score of approximately 0.8846 indicates that 17 pounds is actually within the range of typical weights for babies less than 2 months old. Thus, while it is on the higher side, it is not statistically unusual for a baby to weigh more than 17 pounds.
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