Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
To find the probability that a randomly selected value from a normal distribution is between 349.9 and 410.4, given the mean (213.8) and standard deviation (75.6), we will follow these steps:
### Step 1: Calculate the Z-scores
First, we need to standardize the values 349.9 and 410.4 by converting them to Z-scores. The Z-score for a value [tex]\( X \)[/tex] in a normal distribution can be calculated using the formula:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
where [tex]\(\mu\)[/tex] is the mean and [tex]\(\sigma\)[/tex] is the standard deviation.
#### Z-score for the lower bound (349.9)
[tex]\[ Z_{\text{lower}} = \frac{349.9 - 213.8}{75.6} \][/tex]
[tex]\[ Z_{\text{lower}} \approx 1.800 \][/tex]
#### Z-score for the upper bound (410.4)
[tex]\[ Z_{\text{upper}} = \frac{410.4 - 213.8}{75.6} \][/tex]
[tex]\[ Z_{\text{upper}} \approx 2.601 \][/tex]
### Step 2: Use the Cumulative Distribution Function (CDF)
Next, we use the cumulative distribution function of the standard normal distribution to find the probabilities corresponding to these Z-scores.
Let [tex]\( \Phi(Z) \)[/tex] denote the CDF of the standard normal distribution.
#### Probability corresponding to [tex]\( Z_{\text{lower}} \)[/tex]
[tex]\[ \Phi(1.800) \approx 0.964 \][/tex]
#### Probability corresponding to [tex]\( Z_{\text{upper}} \)[/tex]
[tex]\[ \Phi(2.601) \approx 0.995 \][/tex]
### Step 3: Find the Probability Between the Two Z-scores
To find the probability that a value lies between our two Z-scores, we subtract the smaller CDF value from the larger CDF value:
[tex]\[ P(1.800 < Z < 2.601) = \Phi(2.601) - \Phi(1.800) \][/tex]
[tex]\[ P(1.800 < Z < 2.601) \approx 0.995 - 0.964 \][/tex]
[tex]\[ P(1.800 < Z < 2.601) \approx 0.031 \][/tex]
Thus, the probability that a randomly selected value from the normal distribution with a mean of 213.8 and a standard deviation of 75.6 is between 349.9 and 410.4 is approximately [tex]\( 0.031 \)[/tex] or [tex]\( 3.125\% \)[/tex].
Therefore,
[tex]\[ P(349.9 < X < 410.4) \approx 0.031 \][/tex]
### Step 1: Calculate the Z-scores
First, we need to standardize the values 349.9 and 410.4 by converting them to Z-scores. The Z-score for a value [tex]\( X \)[/tex] in a normal distribution can be calculated using the formula:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
where [tex]\(\mu\)[/tex] is the mean and [tex]\(\sigma\)[/tex] is the standard deviation.
#### Z-score for the lower bound (349.9)
[tex]\[ Z_{\text{lower}} = \frac{349.9 - 213.8}{75.6} \][/tex]
[tex]\[ Z_{\text{lower}} \approx 1.800 \][/tex]
#### Z-score for the upper bound (410.4)
[tex]\[ Z_{\text{upper}} = \frac{410.4 - 213.8}{75.6} \][/tex]
[tex]\[ Z_{\text{upper}} \approx 2.601 \][/tex]
### Step 2: Use the Cumulative Distribution Function (CDF)
Next, we use the cumulative distribution function of the standard normal distribution to find the probabilities corresponding to these Z-scores.
Let [tex]\( \Phi(Z) \)[/tex] denote the CDF of the standard normal distribution.
#### Probability corresponding to [tex]\( Z_{\text{lower}} \)[/tex]
[tex]\[ \Phi(1.800) \approx 0.964 \][/tex]
#### Probability corresponding to [tex]\( Z_{\text{upper}} \)[/tex]
[tex]\[ \Phi(2.601) \approx 0.995 \][/tex]
### Step 3: Find the Probability Between the Two Z-scores
To find the probability that a value lies between our two Z-scores, we subtract the smaller CDF value from the larger CDF value:
[tex]\[ P(1.800 < Z < 2.601) = \Phi(2.601) - \Phi(1.800) \][/tex]
[tex]\[ P(1.800 < Z < 2.601) \approx 0.995 - 0.964 \][/tex]
[tex]\[ P(1.800 < Z < 2.601) \approx 0.031 \][/tex]
Thus, the probability that a randomly selected value from the normal distribution with a mean of 213.8 and a standard deviation of 75.6 is between 349.9 and 410.4 is approximately [tex]\( 0.031 \)[/tex] or [tex]\( 3.125\% \)[/tex].
Therefore,
[tex]\[ P(349.9 < X < 410.4) \approx 0.031 \][/tex]
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.