Answer :
Certainly! Let's solve the problem step-by-step.
### Step 1: Understand the problem
We need to find the probability that a normally distributed random variable [tex]\( x \)[/tex] with a mean [tex]\( \mu = 500 \)[/tex] and a standard deviation [tex]\( \sigma = 150 \)[/tex] is less than 427. This can be expressed as [tex]\( P(x < 427) \)[/tex].
### Step 2: Calculate the z-score
The z-score is a way to measure how many standard deviations [tex]\( x \)[/tex] is away from the mean. The formula to calculate the z-score is:
[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]
For this problem:
- [tex]\( x = 427 \)[/tex]
- [tex]\( \mu = 500 \)[/tex]
- [tex]\( \sigma = 150 \)[/tex]
Plugging the values into the formula:
[tex]\[ z = \frac{427 - 500}{150} \][/tex]
[tex]\[ z = \frac{-73}{150} \][/tex]
[tex]\[ z \approx -0.4867 \][/tex]
### Step 3: Use the z-score to find the probability
Next, we need to find the cumulative probability corresponding to the z-score. The cumulative distribution function (CDF) for a standard normal distribution (mean 0, standard deviation 1) gives the probability that a standard normal variable is less than or equal to a specific value.
Using standard normal distribution tables or a calculator:
- Look up [tex]\( z = -0.4867 \)[/tex] in the cumulative distribution function table, or use a function for the standard normal distribution CDF.
The value of the cumulative probability for [tex]\( z \approx -0.4867 \)[/tex] is approximately 0.3138.
### Step 4: Round the answer
The final step is to round the result to the nearest ten-thousandth place.
So, the probability [tex]\( P(x < 427) \)[/tex] is approximately [tex]\( 0.3138 \)[/tex].
### Conclusion
[tex]\[ \boxed{0.3138} \][/tex]
This is the probability that a normally distributed random variable [tex]\( x \)[/tex] with a mean of 500 and a standard deviation of 150 is less than 427.
### Step 1: Understand the problem
We need to find the probability that a normally distributed random variable [tex]\( x \)[/tex] with a mean [tex]\( \mu = 500 \)[/tex] and a standard deviation [tex]\( \sigma = 150 \)[/tex] is less than 427. This can be expressed as [tex]\( P(x < 427) \)[/tex].
### Step 2: Calculate the z-score
The z-score is a way to measure how many standard deviations [tex]\( x \)[/tex] is away from the mean. The formula to calculate the z-score is:
[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]
For this problem:
- [tex]\( x = 427 \)[/tex]
- [tex]\( \mu = 500 \)[/tex]
- [tex]\( \sigma = 150 \)[/tex]
Plugging the values into the formula:
[tex]\[ z = \frac{427 - 500}{150} \][/tex]
[tex]\[ z = \frac{-73}{150} \][/tex]
[tex]\[ z \approx -0.4867 \][/tex]
### Step 3: Use the z-score to find the probability
Next, we need to find the cumulative probability corresponding to the z-score. The cumulative distribution function (CDF) for a standard normal distribution (mean 0, standard deviation 1) gives the probability that a standard normal variable is less than or equal to a specific value.
Using standard normal distribution tables or a calculator:
- Look up [tex]\( z = -0.4867 \)[/tex] in the cumulative distribution function table, or use a function for the standard normal distribution CDF.
The value of the cumulative probability for [tex]\( z \approx -0.4867 \)[/tex] is approximately 0.3138.
### Step 4: Round the answer
The final step is to round the result to the nearest ten-thousandth place.
So, the probability [tex]\( P(x < 427) \)[/tex] is approximately [tex]\( 0.3138 \)[/tex].
### Conclusion
[tex]\[ \boxed{0.3138} \][/tex]
This is the probability that a normally distributed random variable [tex]\( x \)[/tex] with a mean of 500 and a standard deviation of 150 is less than 427.
