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    Q1. An ordinary dice was thrown 50 times and resulting scores were summarised in a frequency table. The mean score was calculated to be 3.42. It was later found that the frequencies 12 and 9, of two consecutive scores, had been swapped. What is the correct value of the mean?

    Q2. Each of the 34 children in a Year 3 class were given a task to perform. The times taken in minutes, correct to the nearest quarter of a minute, were as follows.

    4, 3+3/4, 5, 6+1/4, 7, 3, 7, 5+1/4, 7+1/2, 8+3/4, 9+1/2, 4+1/2
    6+1/2, 4+1/4, 8, 7+1/4, 6+3/4, 5+3/4, 4+3/4, 8+1/4, 7, 3+1/2, 5+1/2, 7+3/48+1/2, 6+1/2, 5, 7+1/4, 6+3/4, 7+3/4, 5+3/4, 6, 7+3/4, 6+1/2
    Form a grouped frequency table with 6 equal classes beginning with with 3 - 3+3/4

    Q3. The lengths of 250 electronic components were measured very accurately. The results are summarised in the following table:
    Length (cm) : <7, 7-7.05, 7.05-7.10, 7.10-7.15, 7.15-7.20, >7.20
    Frequency: 10, 63, 77, 65, 30, 5
    Given that 10% of the computers are scrapped because they are too short and 8% are scrapped because they are too long, use a cumulative frequency diagram to estimate limits for the length of an acceptable component.

    Q4. A bag contains 4 £1 coins and 12 similar foreign coins. Sophie selects one coin at random from the bag. She keeps the coin if it is a £1 coin, otherwise she returns it to the bag and selects again.

    (a) Find the expected number of selections that Sophie makes, up to and including the first £1 coin.
    (b) Given that Sophie picks a £1 coin on her first selection, find the expected number of extra selections that Sophie makes to obtain a second £1 coin.

    Q5. The lengths of 120 nails of nominal length 3 cm were measured, each correct to the nearest 0.05 cm. The results are summarised in the following table.
    Length (cm): 2.85 2.9 2.95 3 3.05 3.10 3.15
    Frequency: 1 11 27 41 26 12 2
    It is claimed that for a roughly symmetrical distribution the statistic obtained by dividing the interquartile range by the standard deviation is approximately 1.3. Calculate the value of this statistic for these data, and comment.

    Q6. Plates of a certain design are painted by a particular factory employee. At the end of each day the plates are inspected and some are rejected. The table shows the number of plates rejected over a period of 30 days.
    #. of rejects: 0 1 2 3 4 5 6
    #. of days: 18 5 3 1 1 1 1
    Show that the standard deviation of the daily number of rejects is approximately equal to one quarter of the range.

    Most of these questions involve some sort of table, the reason for that is well I'm really awful at table questions so if you explain the beginning or give some sort of tip that should be enough... as for Q1 and Q4, I am completely lost
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    (Original post by verify360)
    Q1. An ordinary dice was thrown 50 times and resulting scores were summarised in a frequency table. The mean score was calculated to be 3.42. It was later found that the frequencies 12 and 9, of two consecutive scores, had been swapped. What is the correct value of the mean?

    Q4. A bag contains 4 £1 coins and 12 similar foreign coins. Sophie selects one coin at random from the bag. She keeps the coin if it is a £1 coin, otherwise she returns it to the bag and selects again.

    (a) Find the expected number of selections that Sophie makes, up to and including the first £1 coin.
    (b) Given that Sophie picks a £1 coin on her first selection, find the expected number of extra selections that Sophie makes to obtain a second £1 coin.

    Most of these questions involve some sort of table, the reason for that is well I'm really awful at table questions so if you explain the beginning or give some sort of tip that should be enough... as for Q1 and Q4, I am completely lost
    We can't do all of you homework without some evidence that you've had a go yourself but let's give you a hand getting started with 1 and 4.

    Q1. You've been given a mean score (3.42) over 50 rolls of a die. This means that if f1, f2, f3, f4, f5, and f6 were the frequencies of 1, 2,...,6 appearing then

    1 x f1 + 2 x f2 + 3 x f3 +...+ 6 x f6 = 50 x 3.41 = 171.

    Now imaging that it was f4 and f5 that were swapped. Instead of 4 x 12 + 5 x 9 appearing in our total, we actually got 4 x 9 + 5 x 12. What is the difference between them? Now imagine it was f3 and f4 that got swapped. Instead of 3 x 12 + 4 x 9, we got 3 x 9 + 4 x 12. What's the difference this time? What do you notice? Now finish the question off.

    Q4. (a) Have you heard of the geometric distribution? If not, look it up; this is what you want to use here.

    (b) This situation is exactly the same as the first, except for there being one less head in the bag.
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    (Original post by verify360)

    Q5. The lengths of 120 nails of nominal length 3 cm were measured, each correct to the nearest 0.05 cm. The results are summarised in the following table.
    Length (cm): 2.85 2.9 2.95 3 3.05 3.10 3.15
    Frequency: 1 11 27 41 26 12 2
    It is claimed that for a roughly symmetrical distribution the statistic obtained by dividing the interquartile range by the standard deviation is approximately 1.3. Calculate the value of this statistic for these data, and comment.

    Q6. Plates of a certain design are painted by a particular factory employee. At the end of each day the plates are inspected and some are rejected. The table shows the number of plates rejected over a period of 30 days.
    #. of rejects: 0 1 2 3 4 5 6
    #. of days: 18 5 3 1 1 1 1
    Show that the standard deviation of the daily number of rejects is approximately equal to one quarter of the range.
    As above, we're not going to do your homework for you, but I'll give out a few tips on these ones. Do you know the formula for the standard deviation? Or for the variance?

    You'll need to start off by finding the sum of (fX) that is the sum of the frequencies multiplied by the occurences, so in the case of Q6:

    sum(fX) = (0 * 18) + (1 * 5) + (2 * 3) + ... + (6 * 1)

    sum(f) = 18 + 5 +3 + ... + 1

    sum (fX^2) = (0^2 * 18) + (1^2 * 5) + (2^2 * 3) + ... (6^2 * 1)

    then you have that \sigma^2 = \frac{1}{\sum f} \left(\sum fX^2 - \frac{\left(\sum fX\right)^2}{\sum f}\right)

    and you plug in the relevant bits and get \sigma^2 the variance, you square root this value to get the standard deviation \sigma
 
 
 
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