Answer :
To find the probability that a randomly selected student has a score between 350 and 550 on a standardized test normally distributed with a mean of 500 and a standard deviation of 110, follow these steps:
1. Calculate the z-scores:
- The z-score represents the number of standard deviations a value is away from the mean.
- The formula for calculating the z-score is [tex]\( z = \frac{(X - \mu)}{\sigma} \)[/tex] where [tex]\( \mu \)[/tex] is the mean and [tex]\( \sigma \)[/tex] is the standard deviation.
For the lower score (350):
[tex]\[ z_{\text{lower}} = \frac{(350 - 500)}{110} = \frac{-150}{110} \approx -1.364 \][/tex]
For the upper score (550):
[tex]\[ z_{\text{upper}} = \frac{(550 - 500)}{110} = \frac{50}{110} \approx 0.455 \][/tex]
2. Use the z-table to find the corresponding probabilities:
- The z-table provides the cumulative probability for each z-score from the mean to the value.
- Approximate the given z-scores using the closest z-values from the table.
For [tex]\( z_{\text{lower}} \approx -1.364 \)[/tex], the closest given z-score is 0.00:
[tex]\[ P(Z < 0.00) = 0.5000 \][/tex]
For [tex]\( z_{\text{upper}} \approx 0.455 \)[/tex], the closest given z-score is 0.45:
[tex]\[ P(Z < 0.45) = 0.6736 \][/tex]
3. Calculate the probability between the z-scores:
- To find the probability that the score falls between the two z-scores, subtract the cumulative probability of the lower z-score from the cumulative probability of the upper z-score.
[tex]\[ P(350 < X < 550) = P(Z < 0.45) - P(Z < 0.00) \][/tex]
Substitute the values from the z-table:
[tex]\[ P(350 < X < 550) = 0.6736 - 0.5000 = 0.1736 \][/tex]
Therefore, the probability that a randomly selected student scores between 350 and 550 on this standardized test is approximately [tex]\( 0.1736 \)[/tex], or 17.36%.
1. Calculate the z-scores:
- The z-score represents the number of standard deviations a value is away from the mean.
- The formula for calculating the z-score is [tex]\( z = \frac{(X - \mu)}{\sigma} \)[/tex] where [tex]\( \mu \)[/tex] is the mean and [tex]\( \sigma \)[/tex] is the standard deviation.
For the lower score (350):
[tex]\[ z_{\text{lower}} = \frac{(350 - 500)}{110} = \frac{-150}{110} \approx -1.364 \][/tex]
For the upper score (550):
[tex]\[ z_{\text{upper}} = \frac{(550 - 500)}{110} = \frac{50}{110} \approx 0.455 \][/tex]
2. Use the z-table to find the corresponding probabilities:
- The z-table provides the cumulative probability for each z-score from the mean to the value.
- Approximate the given z-scores using the closest z-values from the table.
For [tex]\( z_{\text{lower}} \approx -1.364 \)[/tex], the closest given z-score is 0.00:
[tex]\[ P(Z < 0.00) = 0.5000 \][/tex]
For [tex]\( z_{\text{upper}} \approx 0.455 \)[/tex], the closest given z-score is 0.45:
[tex]\[ P(Z < 0.45) = 0.6736 \][/tex]
3. Calculate the probability between the z-scores:
- To find the probability that the score falls between the two z-scores, subtract the cumulative probability of the lower z-score from the cumulative probability of the upper z-score.
[tex]\[ P(350 < X < 550) = P(Z < 0.45) - P(Z < 0.00) \][/tex]
Substitute the values from the z-table:
[tex]\[ P(350 < X < 550) = 0.6736 - 0.5000 = 0.1736 \][/tex]
Therefore, the probability that a randomly selected student scores between 350 and 550 on this standardized test is approximately [tex]\( 0.1736 \)[/tex], or 17.36%.
