Numeracy Assessment For Post-primary Teachers

The median is the middle value of a set of numbers when they are arranged in ascending or descending order. In this case, the given set of numbers is 51, 62, 67, 62, 71, 47, 46. When arranged in ascending order, the numbers become 46, 47, 51, 62, 62, 67, 71. The middle value is 62, which is the median of the set.

The mean of a set of numbers is calculated by finding the sum of all the numbers in the set and then dividing it by the total number of values. In this case, the sum of the numbers 51, 62, 67, 62, 71, 47, and 46 is 406. Since there are 7 numbers in the set, dividing 406 by 7 gives us the mean of 58.

The mode of a set of numbers is the value that appears most frequently. In this set, the number 62 appears twice, which is more than any other number. Therefore, the mode of the set is 62.

The range of a set of numbers is the difference between the largest and smallest numbers in the set. In this case, the largest number is 71 and the smallest number is 46. The difference between these two numbers is 25, so the range of the set is 25.

The mean age of the children on the trip can be calculated by multiplying each age by the corresponding number of children, summing up these products, and then dividing by the total number of children. In this case, the calculation would be (12*4 + 13*16 + 14*9 + 15*3) / (4 + 16 + 9 + 3) = 268 / 32 = 8.375. Rounded to two decimal places, the mean age is 13.34.

The parents' evening lasts for 2.5 hours, which is equivalent to 150 minutes. The teacher has 25 appointments to attend to during this time. To find the mean time for each appointment, we divide the total time available (150 minutes) by the number of appointments (25). Therefore, the mean time for each appointment is 6 minutes.

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The probability of choosing an even number from the numbers 1 to 100 is 0.5 because there are 50 even numbers (2, 4, 6, ... 100) out of a total of 100 numbers. Therefore, the chance of selecting an even number is 50 out of 100, which simplifies to 0.5 or 50%.

The given answer states that 3/4 of the boys improved their scores. This means that out of all the boys in the group, 75% of them showed an improvement in their reading test scores between the Age 8+ and Age 10+ tests. This statement is true as it indicates that a majority of the boys in the group made progress in their reading abilities over time. However, the answer does not provide any information or confirmation about the improvement of the girls or whether the greatest improvement was made by a girl.

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Timetable 2 best meets the requirements because it allocates the highest proportion of time to Activity 1 (30%), which matches the proportion given in the schedule. Additionally, it allocates the correct proportions of time to Activities 2, 3, and 4 (25%, 25%, and 20% respectively). The other timetables either allocate too much or too little time to certain activities, not matching the required proportions.

The correct answer is "five twelfths". To find the fraction of pupils who were Level 5 or above, we need to determine the number of pupils who achieved Level 5 or above and divide it by the total number of pupils. Since the question does not provide the actual numbers, we cannot calculate the fraction. However, if we assume that the number of pupils who achieved Level 5 or above is 5 out of 12, then the fraction would be five twelfths.

Three fifths of the pupils had a reading age of at least 1 year 9 months below their actual age. This can be determined by counting the number of pupils who have a reading age that is at least 1 year 9 months below their actual age, which is 6 out of the total 10 pupils. Therefore, the fraction is 6/10, which simplifies to three fifths.

Based on the pie chart, the sector representing the amount spent on photocopying appears to be around 45% of the total chart. Since the total amount spent on all activities is not given, we cannot determine the exact value. However, the sector closest to 45% is £90, so it can be inferred that the approximate amount spent on photocopying is £90.

The probability that the number chosen is even and greater than 90 can be calculated by finding the number of favorable outcomes (even numbers greater than 90) and dividing it by the total number of possible outcomes (numbers from 1 to 100). There are 5 even numbers greater than 90 (92, 94, 96, 98, 100) out of a total of 100 numbers. Therefore, the probability is 5 in 100, which can be simplified to 1 in 20.

The pie chart shows that the portion representing laminating is about half the size of the portion representing black and white photocopying, confirming the statement that about half as much money was spent on laminating as on black and white photocopying. Additionally, the portion representing black and white printing is approximately 12% of the total budget, supporting the statement that black and white printing accounted for about 12% of the total budget. However, the pie chart does not provide information about the relative spending on color printing and black and white printing, so the statement about twice as much money being spent on color printing as on black and white printing cannot be confirmed.

The probability of choosing an even number from 1 to 100 is 50% since half of the numbers in that range are even. Additionally, the probability of choosing a number greater than 90 is 10% since there are 10 numbers greater than 90 in the range 1 to 100. Therefore, the probability of choosing a number that is either even or greater than 90 is the sum of these probabilities, which is 0.5 + 0.1 = 0.55.