In a school all First Years sat a common maths exam - Leaving Cert Mathematics - Question 8 - 2021

To find the percentage of students who received the Certificate of Merit, we need to calculate the probability of students scoring between 165 and 210 marks.

First, we find the z-scores for both marks:

For 165 marks: $z = \frac{165 - 176}{36} = \frac{-11}{36} \approx -0.305$

For 210 marks: $z = \frac{210 - 176}{36} = \frac{34}{36} \approx 0.944$

Next, we look up these z-scores in a standard normal distribution table or use a calculator:

• The probability for $z = -0.305$ is approximately 0.3803.
• The probability for $z = 0.944$ is approximately 0.8264.

Now, to find the percentage of students with marks between 165 and 210: $P(165 < x < 210) = P(z < 0.944) - P(z < -0.305) = 0.8264 - 0.3803 = 0.4461$

Thus, approximately 44.61% of First Years received the Certificate of Merit.

To find the test statistic, we will calculate the z-score for the sample mean of 19.8 hours.

Using the formula: $z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ Where:

• $\bar{x} = 19.8$ (sample mean)
• $\mu = 21$ (population mean)
• $\sigma = 5.2$ (population standard deviation)
• $n = 60$ (sample size)

Now substituting the values: $z = \frac{19.8 - 21}{\frac{5.2}{\sqrt{60}}} \approx \frac{-1.2}{0.6722} \approx -1.787$

Thus, the test statistic (z-score) is approximately -1.787.

To find the p-value associated with the z-score of -1.787, we look it up in the z-table:

A z-score of -1.787 corresponds to a left-tailed probability of approximately 0.0372. Since we are performing a two-tailed test:

$p = 2 imes P(Z < -1.787) = 2 imes 0.0372 = 0.0744$

Since the p-value (0.0744) is greater than the significance level of 0.05, we fail to reject the null hypothesis.

Conclusion: There is not enough evidence to say that the claim in the news report is incorrect.

To find this probability, we use the concept of conditional probability. The first three keys drawn must be from the other categories (classroom or science).

The probability of drawing 3 non-office keys in succession is:

• First draw: rac{18}{23} (18 non-office keys)
• Second draw: rac{17}{22} (17 non-office keys left)
• Third draw: rac{16}{21} (16 non-office keys left)
• Fourth draw must be an office key: rac{5}{20} (5 office keys left)

Thus, the overall probability is: $P = \left(\frac{18}{23}\right) \times \left(\frac{17}{22}\right) \times \left(\frac{16}{21}\right) \times \left(\frac{5}{20}\right) \approx 0.1152$

Therefore, the probability that the 4th key drawn is the first office key drawn is approximately 0.1152.

For this question, since we are drawing 3 keys without replacement, we will consider all the combinations:

1. General Classroom, Science Lab, Office
2. General Classroom, Office, Science Lab
3. Science Lab, General Classroom, Office
4. Science Lab, Office, General Classroom
5. Office, General Classroom, Science Lab
6. Office, Science Lab, General Classroom

The probability can thus be calculated as:

• Total ways to draw 3 out of 23: $\binom{23}{3}$
• Total favorable outcomes for each arrangement:
  • Classroom (12): $\binom{12}{1}$
  • Science Lab (6): $\binom{6}{1}$
  • Office (5): $\binom{5}{1}$

Now we calculate: $P = \frac{3! \times \binom{12}{1} \times \binom{6}{1} \times \binom{5}{1}}{\binom{23}{3}}\approx 0.2033$

Therefore, the probability that one of them is for a general classroom, one is for a science lab, and one is for an office is approximately 0.2033.