A group of students sat an exam - Leaving Cert Mathematics - Question 4 - 2022

Using the empirical rule:

1. Identify the Range:

   • Mean = 100
   • Standard Deviation = 20

2. Calculate Range for 80 to 120:

   • 80 lies 1 standard deviation below the mean.
   • 120 lies 1 standard deviation above the mean.

3. Percentage within Range:

   • According to the empirical rule, approximately 68% of the data falls within 1 standard deviation from the mean.

Thus, the percentage of people who scored between 80 and 120 is 68%.

1. Calculate the Z-Score:

   • For the top 2.5%, find the corresponding Z-score, which is approximately 1.96.

2. Calculate the Score:

   • Use the following formula: $X = ext{mean} + (Z imes ext{standard deviation})$
   • Plug in values: $X = 100 + (1.96 imes 20) = 100 + 39.2 = 139.2$

Thus, the least score needed to achieve the grade of ‘Exceptional’ is 140.

1. Calculate Range:

   • Range = Maximum - Minimum = 113 - 82 = 31.

2. Calculate Standard Deviation:

   • Mean = (104 + 82 + 94 + 113 + 98 + 105) / 6 = 99.
   • Standard Deviation formula:
     $S.D. = \sqrt{\frac{(x_1 - \overline{x})^2 + (x_2 - \overline{x})^2 + ... + (x_n - \overline{x})^2}{n}}$
   • Calculate:
     $S.D. = \sqrt{\frac{(104-99)^2 + (82-99)^2 + (94-99)^2 + (113-99)^2 + (98-99)^2 + (105-99)^2}{6}}$
   • This yields:
     $S.D. = 9.758 ext{ (approximately 9.8 to 1 decimal place)}$

Thus, the standard deviation is approximately 9.8.