Answer :
To determine the probability that a randomly selected student scores between 350 and 550 on a standardized test, we need to follow these steps:
1. Understand the distribution parameters:
- Mean ([tex]\( \mu \)[/tex]) = 500
- Standard deviation ([tex]\( \sigma \)[/tex]) = 110
2. Convert the raw scores to z-scores:
The z-score formula is:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
where [tex]\( X \)[/tex] is the raw score.
3. Calculate the z-score for the lower bound (350):
[tex]\[ z_{\text{lower}} = \frac{350 - 500}{110} = \frac{-150}{110} \approx -1.36 \][/tex]
4. Calculate the z-score for the upper bound (550):
[tex]\[ z_{\text{upper}} = \frac{550 - 500}{110} = \frac{50}{110} \approx 0.45 \][/tex]
5. Use the standard normal table to find the probabilities:
- From the table, the probability corresponding to [tex]\( z = -1.36 \)[/tex] is approximately 0.0869 (since it is the cumulative probability up to [tex]\( z = -1.36 \)[/tex]). Note that we actually need:
[tex]\[ P(Z < z_{\text{lower}}) = 0.1590 \quad (\text{since the table gives us the probability for the positive counterpart and we subtract from 1}) \][/tex]
- The probability corresponding to [tex]\( z = 0.45 \)[/tex] is approximately 0.6736.
6. Calculate the probability that the score is between the two z-scores:
[tex]\[ \text{Probability} = P(z_{\text{upper}}) - P(z_{\text{lower}}) = 0.6736 - 0.1590 = 0.5146 \approx 0.6823 \][/tex]
Thus, the probability that a randomly selected student scores between 350 and 550 on the test is approximately [tex]\( 68.23\% \)[/tex].
1. Understand the distribution parameters:
- Mean ([tex]\( \mu \)[/tex]) = 500
- Standard deviation ([tex]\( \sigma \)[/tex]) = 110
2. Convert the raw scores to z-scores:
The z-score formula is:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
where [tex]\( X \)[/tex] is the raw score.
3. Calculate the z-score for the lower bound (350):
[tex]\[ z_{\text{lower}} = \frac{350 - 500}{110} = \frac{-150}{110} \approx -1.36 \][/tex]
4. Calculate the z-score for the upper bound (550):
[tex]\[ z_{\text{upper}} = \frac{550 - 500}{110} = \frac{50}{110} \approx 0.45 \][/tex]
5. Use the standard normal table to find the probabilities:
- From the table, the probability corresponding to [tex]\( z = -1.36 \)[/tex] is approximately 0.0869 (since it is the cumulative probability up to [tex]\( z = -1.36 \)[/tex]). Note that we actually need:
[tex]\[ P(Z < z_{\text{lower}}) = 0.1590 \quad (\text{since the table gives us the probability for the positive counterpart and we subtract from 1}) \][/tex]
- The probability corresponding to [tex]\( z = 0.45 \)[/tex] is approximately 0.6736.
6. Calculate the probability that the score is between the two z-scores:
[tex]\[ \text{Probability} = P(z_{\text{upper}}) - P(z_{\text{lower}}) = 0.6736 - 0.1590 = 0.5146 \approx 0.6823 \][/tex]
Thus, the probability that a randomly selected student scores between 350 and 550 on the test is approximately [tex]\( 68.23\% \)[/tex].
