The level of cholesterol in the blood for all men aged 20 to 34 follows a Normal distribution when mean $$ \mu_{M}=188 $$ milligrams © BFW Publishers per deciliter ( $$ \mathrm{mg} / \mathrm{dl} $$ ) and standard deviation $$ \sigma_{M}=41 \mathrm{mg} / \mathrm{dl} $$. For 14 -year-old boys, blood cholesterol levels follow a Normal distribution with mean $$ \mu_{B}=170 \mathrm{mg} / \mathrm{dl} $$ and standard deviation $$ \sigma_{B}=30 \mathrm{mg} / \mathrm{dl} $$. Suppose we select independent SRSs of 25 men aged 20 to 34 and 36 boys aged 14 and calculate the sample mean cholesterol levels $$ \bar{x}_{M} $$ and $$ \bar{x}_{B} $$. What is the probability of getting a difference in sample means $$ \bar{x}_{M}-\bar{x}_{B} $$ that's less than $$ 0 \mathrm{mg} / \mathrm{dl} $$ ? Round to 4 decimal places.

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To find the probability of getting a difference in sample means $$ \bar{x}_{M} - \bar{x}_{B} < 0 \, \text{mg/dl} $$, we can follow these steps:

1. **Identify the means and standard deviations**:
   - For men aged 20 to 34:
     - Mean: $$ \mu_{M} = 188 \, \text{mg/dl} $$
     - Standard deviation: $$ \sigma_{M} = 41 \, \text{mg/dl} $$
     - Sample size: $$ n_{M} = 25 $$

- For boys aged 14:
     - Mean: $$ \mu_{B} = 170 \, \text{mg/dl} $$
     - Standard deviation: $$ \sigma_{B} = 30 \, \text{mg/dl} $$
     - Sample size: $$ n_{B} = 36 $$

2. **Calculate the standard error of the difference in sample means**:
   The standard error (SE) of the difference between two independent sample means is given by:
   $$
   SE = \sqrt{\frac{\sigma_{M}^2}{n_{M}} + \frac{\sigma_{B}^2}{n_{B}}}
   $$

3. **Calculate the mean of the difference in sample means**:
   The mean of the difference $$ \mu_{\bar{x}_{M} - \bar{x}_{B}} $$ is:
   $$
   \mu_{\bar{x}_{M} - \bar{x}_{B}} = \mu_{M} - \mu_{B}
   $$

4. **Standardize the difference**:
   We want to find $$ P(\bar{x}_{M} - \bar{x}_{B} < 0) $$. This can be standardized using the Z-score:
   $$
   Z = \frac{(\bar{x}_{M} - \bar{x}_{B}) - \mu_{\bar{x}_{M} - \bar{x}_{B}}}{SE}
   $$

5. **Calculate the probability**:
   We will find $$ P(Z < z) $$ where $$ z $$ is the standardized value corresponding to $$ \bar{x}_{M} - \bar{x}_{B} = 0 $$.

Now, let's perform the calculations step by step.

### Step 2: Calculate the Standard Error
$$
SE = \sqrt{\frac{41^2}{25} + \frac{30^2}{36}}
$$

### Step 3: Calculate the Mean of the Difference
$$
\mu_{\bar{x}_{M} - \bar{x}_{B}} = 188 - 170
$$

### Step 4: Standardize the Difference
We will find $$ Z $$ when $$ \bar{x}_{M} - \bar{x}_{B} = 0 $$:
$$
Z = \frac{0 - \mu_{\bar{x}_{M} - \bar{x}_{B}}}{SE}
$$

### Step 5: Calculate the Probability
Finally, we will find $$ P(Z < z) $$.

Let's perform these calculations.
Calculate the value by following steps:
- step0: Calculate:
  $$188-170$$
- step1: Subtract the numbers:
  $$18$$
Calculate or simplify the expression $$ \sqrt((41^2)/25 + (30^2)/36) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\sqrt{\frac{41^{2}}{25}+\frac{30^{2}}{36}}$$
- step1: Divide the terms:
  $$\sqrt{\frac{41^{2}}{25}+5^{2}}$$
- step2: Add the numbers:
  $$\sqrt{\frac{2306}{25}}$$
- step3: Use the properties of radicals:
  $$\frac{\sqrt{2306}}{\sqrt{25}}$$
- step4: Simplify the expression:
  $$\frac{\sqrt{2306}}{5}$$
Calculate or simplify the expression $$ (0 - 18) / 9.604166 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(0-18\right)}{9.604166}$$
- step1: Remove the parentheses:
  $$\frac{0-18}{9.604166}$$
- step2: Remove 0:
  $$\frac{-18}{9.604166}$$
- step3: Convert the expressions:
  $$\frac{-18}{\frac{4802083}{500000}}$$
- step4: Multiply by the reciprocal:
  $$-18\times \frac{500000}{4802083}$$
- step5: Multiply:
  $$-\frac{18\times 500000}{4802083}$$
- step6: Multiply:
  $$-\frac{9000000}{4802083}$$
Let's summarize the calculations we have performed so far:

1. **Standard Error (SE)**:
   $$
   SE \approx 9.6042 \, \text{mg/dl}
   $$

2. **Mean of the Difference**:
   $$
   \mu_{\bar{x}_{M} - \bar{x}_{B}} = 188 - 170 = 18 \, \text{mg/dl}
   $$

3. **Standardized Value (Z)**:
   $$
   Z = \frac{0 - 18}{9.6042} \approx -1.8742
   $$

Now, we need to find the probability $$ P(Z < -1.8742) $$. This can be done using the standard normal distribution table or a calculator.

To find this probability, we can use the cumulative distribution function (CDF) for the standard normal distribution.

Let's calculate $$ P(Z < -1.8742) $$.

Using a standard normal distribution table or calculator, we find:

$$
P(Z < -1.8742) \approx 0.0301
$$

Thus, the probability of getting a difference in sample means $$ \bar{x}_{M} - \bar{x}_{B} < 0 \, \text{mg/dl} $$ is approximately:

$$
\boxed{0.0301}
$$

This value is rounded to four decimal places.
The probability of the difference in sample means being less than 0 mg/dl is approximately 0.0301.

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