Answer :
To solve this problem, we need to find the percentage of scores that fall between 75.8 and 89 on a normally distributed high school entrance exam, given a mean ([tex]\(\mu\)[/tex]) of 82.4 and a standard deviation ([tex]\(\sigma\)[/tex]) of 3.3.
Step-by-step solution:
1. Find the Z-scores:
- The Z-score tells us how many standard deviations away a particular value is from the mean. We use the formula:
[tex]\[
Z = \frac{X - \mu}{\sigma}
\][/tex]
- For the lower bound (75.8):
[tex]\[
Z_{\text{lower}} = \frac{75.8 - 82.4}{3.3} \approx -2.0
\][/tex]
- For the upper bound (89):
[tex]\[
Z_{\text{upper}} = \frac{89 - 82.4}{3.3} \approx 2.0
\][/tex]
2. Use the standard normal distribution table:
- With the Z-scores calculated, we look up these values in the standard normal distribution (Z-table) to find the cumulative probability.
- The cumulative probability for [tex]\(Z_{\text{lower}} \approx -2.0\)[/tex] is around 0.0228 (or 2.28%).
- The cumulative probability for [tex]\(Z_{\text{upper}} \approx 2.0\)[/tex] is around 0.9772 (or 97.72%).
3. Calculate the probability between the Z-scores:
- Subtract the cumulative probability of the lower bound from the cumulative probability of the upper bound:
[tex]\[
\text{Probability} = 0.9772 - 0.0228 = 0.9544
\][/tex]
4. Convert the probability to a percentage:
- Convert the probability to a percentage by multiplying by 100:
[tex]\[
0.9544 \times 100 \approx 95.4\%
\][/tex]
Hence, approximately 95.4% of the scores lie between 75.8 and 89. This matches the given option of 95%, so the percentage of scores between 75.8 and 89 is 95%.
Step-by-step solution:
1. Find the Z-scores:
- The Z-score tells us how many standard deviations away a particular value is from the mean. We use the formula:
[tex]\[
Z = \frac{X - \mu}{\sigma}
\][/tex]
- For the lower bound (75.8):
[tex]\[
Z_{\text{lower}} = \frac{75.8 - 82.4}{3.3} \approx -2.0
\][/tex]
- For the upper bound (89):
[tex]\[
Z_{\text{upper}} = \frac{89 - 82.4}{3.3} \approx 2.0
\][/tex]
2. Use the standard normal distribution table:
- With the Z-scores calculated, we look up these values in the standard normal distribution (Z-table) to find the cumulative probability.
- The cumulative probability for [tex]\(Z_{\text{lower}} \approx -2.0\)[/tex] is around 0.0228 (or 2.28%).
- The cumulative probability for [tex]\(Z_{\text{upper}} \approx 2.0\)[/tex] is around 0.9772 (or 97.72%).
3. Calculate the probability between the Z-scores:
- Subtract the cumulative probability of the lower bound from the cumulative probability of the upper bound:
[tex]\[
\text{Probability} = 0.9772 - 0.0228 = 0.9544
\][/tex]
4. Convert the probability to a percentage:
- Convert the probability to a percentage by multiplying by 100:
[tex]\[
0.9544 \times 100 \approx 95.4\%
\][/tex]
Hence, approximately 95.4% of the scores lie between 75.8 and 89. This matches the given option of 95%, so the percentage of scores between 75.8 and 89 is 95%.
