The diagram shows the standard normal curve - Leaving Cert Mathematics - Question 2 - 2018

To compare Mary’s performance, we will use the z-score formula:

z = rac{X - ext{mean}}{ ext{standard deviation}}

For Maths:

• Mean = 70
• Standard deviation = 15
• Mary’s score = 65

Calculating Mary's z-score in Maths:

z_{Maths} = rac{65 - 70}{15} = -rac{1}{3}

For English:

• Mean = 72
• Standard deviation = 10
• Mary’s score = 68

Calculating Mary's z-score in English:

z_{English} = rac{68 - 72}{10} = -rac{4}{10} = -0.4

Since $z_{Maths} < z_{English}$, it indicates that Mary did better in Maths relative to her peers.

To find the least whole number mark for the top 15%, we need to find the z-score for 15%:

$P(Z < z) = 0.85$

From the Z-table, this corresponds to:

$z ext{ is } 1.04$

Using the z-score formula, we can solve for the raw score (x):

z = rac{x - ext{mean}}{ ext{standard deviation}}

Replacing the known values:

1.04 = rac{x - 72}{10}

Solving for $x$:

$x - 72 = 10 * 1.04$ $x = 72 + 10.4 = 82.4$

Thus, the least whole number mark is: 83.

To estimate the percentage of students who scored between 52 and 82, we will calculate the z-scores for both values:

1. For 52:

z_{52} = rac{52 - 72}{10} = -2

2. For 82:

z_{82} = rac{82 - 72}{10} = 1

Using the empirical rule:

• Approximately 68% of the data lies within 1 standard deviation.

From the Z-table:

• $P(Z < 1) ≈ 0.8413$
• $P(Z < -2) ≈ 0.0228$

Thus, the percentage of students scoring between 52 and 82 is:

$P(52 < X < 82) = P(Z < 1) - P(Z < -2)$

Calculating this gives:

$0.8413 - 0.0228 = 0.8185$

Therefore, the estimated percentage is: 81.85%.