A population of values has a normal distribution with $$ \mu=84.3 $$ and $$ \sigma=82.9 $$. You intend to draw a random sample of size $$ n=201 $$. Find the probability that a single randomly selected value is between 79.6 and 87.8 . $$ P(79.6

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Alright, I need to solve this probability problem involving a normal distribution. Let's break it down step by step to ensure I cover all aspects thoroughly.

**Given:**
- Population mean, $$ \mu = 84.3 $$
- Population standard deviation, $$ \sigma = 82.9 $$
- Sample size, $$ n = 201 $$
- We need to find:
  1. $$ P(79.6 < X < 87.8) $$ for a single randomly selected value.
  2. $$ P(79.6 < M < 87.8) $$ for the sample mean of size $$ n = 201 $$.

**Understanding the Problem:**
First, I need to recall that for a normal distribution, the probability of a value falling within a certain range can be found using the standard normal distribution (Z-distribution). The Z-score formula is essential here:
$$
Z = \frac{X - \mu}{\sigma}
$$
However, when dealing with sample means, the standard error (SE) comes into play:
$$
SE = \frac{\sigma}{\sqrt{n}}
$$
And the Z-score for the sample mean is:
$$
Z = \frac{M - \mu}{SE}
$$

**Calculating the Z-scores:**
1. **For a single value ($$ X $$):**
   - $$ Z_1 = \frac{79.6 - 84.3}{82.9} $$
   - $$ Z_2 = \frac{87.8 - 84.3}{82.9} $$
   
2. **For the sample mean ($$ M $$):**
   - $$ SE = \frac{82.9}{\sqrt{201}} $$
   - $$ Z_1 = \frac{79.6 - 84.3}{SE} $$
   - $$ Z_2 = \frac{87.8 - 84.3}{SE} $$

**Calculations:**
1. **Single Value:**
   - $$ Z_1 = \frac{79.6 - 84.3}{82.9} = \frac{-4.7}{82.9} \approx -0.0567 $$
   - $$ Z_2 = \frac{87.8 - 84.3}{82.9} = \frac{3.5}{82.9} \approx 0.0422 $$
   
2. **Sample Mean:**
   - $$ SE = \frac{82.9}{\sqrt{201}} \approx \frac{82.9}{14.177} \approx 5.85 $$
   - $$ Z_1 = \frac{79.6 - 84.3}{5.85} = \frac{-4.7}{5.85} \approx -0.803 $$
   - $$ Z_2 = \frac{87.8 - 84.3}{5.85} = \frac{3.5}{5.85} \approx 0.598 $$

**Using Z-tables or Calculators:**
Now, I'll use standard normal distribution tables or a calculator to find the probabilities corresponding to these Z-scores.

1. **Single Value:**
   - $$ P(-0.0567 < Z < 0.0422) $$
   - Looking up $$ Z = 0.0422 $$ gives approximately 0.5170
   - Looking up $$ Z = -0.0567 $$ gives approximately 0.4772
   - Therefore, $$ P(79.6 < X < 87.8) = 0.5170 - 0.4772 = 0.0398 $$

2. **Sample Mean:**
   - $$ P(-0.803 < Z < 0.598) $$
   - Looking up $$ Z = 0.598 $$ gives approximately 0.7236
   - Looking up $$ Z = -0.803 $$ gives approximately 0.2119
   - Therefore, $$ P(79.6 < M < 87.8) = 0.7236 - 0.2119 = 0.5117 $$

**Double-Checking:**
It's always good to verify the calculations to ensure accuracy.

1. **Single Value:**
   - Recalculating $$ Z $$ values:
     - $$ Z_1 = -0.0567 $$
     - $$ Z_2 = 0.0422 $$
   - Probabilities:
     - $$ P(Z < 0.0422) \approx 0.5170 $$
     - $$ P(Z < -0.0567) \approx 0.4772 $$
   - Difference: $$ 0.5170 - 0.4772 = 0.0398 $$

2. **Sample Mean:**
   - Recalculating $$ SE $$:
     - $$ SE = 5.85 $$
   - $$ Z $$ values:
     - $$ Z_1 = -0.803 $$
     - $$ Z_2 = 0.598 $$
   - Probabilities:
     - $$ P(Z < 0.598) \approx 0.7236 $$
     - $$ P(Z < -0.803) \approx 0.2119 $$
   - Difference: $$ 0.7236 - 0.2119 = 0.5117 $$

**Conclusion:**
After performing the calculations and verifying them, the probabilities are as follows:

1. $$ P(79.6 < X < 87.8) = 0.0398 $$
2. $$ P(79.6 < M < 87.8) = 0.5117 $$

These probabilities are accurate to four decimal places, as required.

$$ P(79.6 < X < 87.8) = 0.0398 $$ $$ P(79.6 < M < 87.8) = 0.5117 $$

A population of values has a normal distribution with \( \mu=84.3 \) and \( \sigma=82.9 \). You intend to draw a random sample of size \( n=201 \). Find the probability that a single randomly selected value is between 79.6 and 87.8 . \( P(79.6

Ask by Harper Hardy. in the United States

Mar 16,2025

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A population of values has a normal distribution with \( \mu=84.3 \) and \( \sigma=82.9 \). You intend to draw a random sample of size \( n=201 \). Find the probability that a single randomly selected value is between 79.6 and 87.8 . \( P(79.6 Ask by Harper Hardy. in the United States Mar 16,2025