Data (4) Flashcards
In a supermarket, two types of chocolate drops were compared.
The weights, a grams, of 20 chocolate drops of brand A are summarised by: Σa = 60.3g, Σa^2 = 219g^2
The mean weight of 30 chocolate drops of brand B was 2.95g, and the standard deviation was 1g
Find the mean weight of a brand A chocolate drop
Mean of a = 60.3/20 = 3.015g
In a supermarket, two types of chocolate drops were compared.
The weights, a grams, of 20 chocolate drops of brand A are summarised by: Σa = 60.3g, Σa^2 = 219g^2
The mean weight of 30 chocolate drops of brand B was 2.95g, and the standard deviation was 1g
Find the standard deviation of the weight of brand A chocolate drops
(S_A)^2 = (219/20) - 3.015^2 = 1.860g^2 S_A = 1.36g
In a supermarket, two types of chocolate drops were compared.
The weights, a grams, of 20 chocolate drops of brand A are summarised by: Σa = 60.3g, Σa^2 = 219g^2
The mean weight of 30 chocolate drops of brand B was 2.95g, and the standard deviation was 1g
Compare the weights of chocolate drops from Brand A and B
Brand A chocolate drops are heavier on average than Brand B. Brand B chocolate drops are generally much closer to their mean weight than Brand A
In a supermarket, two types of chocolate drops were compared.
The weights, a grams, of 20 chocolate drops of brand A are summarised by: Σa = 60.3g, Σa^2 = 219g^2
The mean weight of 30 chocolate drops of brand B was 2.95g, and the standard deviation was 1g
Find the standard deviation of the weight of all 50 chocolate drops
Mean of A and B = (Σa + Σb)/50 = (60.3 + (30 * 2.95))/50 = 2.976g
((Σb^2)/30) - 2.95^2 = 1, so Σb^2 = 291.075
Variance of A and B = ((Σa^2 + Σb^2)/50) - 2.976^2
((219 + 291.075)/50) - 2.976^2 = 1.3449
¬(1.3449) = 1.16g
The profits of 100 businesses are given in the table:
Profit (£p million): 4.5 <= p < 5
Number of businesses: 21
Profit (£p million): 5 <= p < 5.5
Number of businesses: 26
Profit (£p million): 5.5 <= p < 6
Number of businesses: 24
Profit (£p million): 6 <= p < 6.5
Number of businesses: 19
Profit (£p million): 6.5 <= p < 7
Number of businesses: 10
Estimate the mean and standard deviation of this data
Class midpoint as x Number of businesses as f Calculate fx Calculate x^2 Calculate fx^2
Estimated mean: (Σfx / Σf)
Estimated variance: (Σfx^2 / Σf) - (estimated mean)^2
Square root the estimated variance to get the standard deviation
The profits of 100 businesses are given in the table:
Profit (£p million): 4.5 <= p < 5
Number of businesses: 21
Profit (£p million): 5 <= p < 5.5
Number of businesses: 26
Profit (£p million): 5.5 <= p < 6
Number of businesses: 24
Profit (£p million): 6 <= p < 6.5
Number of businesses: 19
Profit (£p million): 6.5 <= p < 7
Number of businesses: 10
Use linear interpolation to estimate the median profit
n / 2 = 50
21 + 26 = 47, so the median is in the 5.5-6 class
Estimated median = 5.5 + 0.5 * (50-47)/24
= £5.5625 million
Find the median and mode number of hits
Total number of people = 38
38 / 2 = 19
Median is the average of the 19th and 20th value, so median = 15.5 hits and mode = 15 hits
An outlier is a data value which is more than 1.5 * (Q_3 - Q_1) above Q_3 or below Q_1
Is 25 an outlier?
LQ = 10th value = 14 UQ = 29th value = 17 IQR = 17 - 14 = 3
And upper fence = 17 + (1.5 * 3) = 21.5
This means that 25 is an outlier
Explain why the median might be considered a more reliable measure of central tendency than the mean for a data set that is thought to contain an outlier
The value of the mean is likely to be affected more than the median by the presence of an outlier
Two workers iron clothes. Each irons ten items, and records the time it takes for each, to the nearest minute
Worker A: 3,5,2,7,10,4,5,5,4,12
Worker B: 3,4,8,6,7,8,9,10,11,9
For Worker A’s times, find: the median and the lower and upper quartile
Times = 2,3,4,4,5,5,5,7,10,12 (n/2) = 5. 5 is a whole number so find the average between the 5th and 6th terms. (5+5)/2 = 5 minutes
(n/4) = 2.5. 2.5 is not a whole number, so the LQ is the 3rd term. Q_1 = 4 minutes
(3n/4) = 7.5. 7.5 is not a whole number, so the LQ is the 8th term. Q_3 = 7 minutes
Two workers iron clothes. Each irons ten items, and records the time it takes for each, to the nearest minute
Worker A: 3,5,2,7,10,4,5,5,4,12
Worker B: 3,4,8,6,7,8,9,10,11,9
Worker A claims he deserves a pay rise because he works faster than worker B. State, giving a reason, whether the data given above supports Worker A’s claim
The data supports Worker A’s claim, as the median for Worker B is higher, so the times for Worker B are generally higher
A group of 19 people played a game. The scores, x, that the people achieved are summarised by:
Σ(x-30) = 228 and Σ(x-30)^2 = 3040
Calculate the mean and the standard deviation of the 19 scores
y = x - 30
Mean of y = 228/19 = 12, so mean of x = mean of y + 30 = 42
Variance of y = (3040/19) - 12^2 = 16
Standard deviation of y = 4
But standard deviation of x = standard deviation of y
So standard deviation of x = 4
A group of 19 people played a game. The scores, x, that the people achieved are summarised by:
Σ(x-30) = 228 and Σ(x-30)^2 = 3040
Show that Σx = 798 and Σx^2 = 33820
Mean of x = (Σx/19) = 42, so Σx = 42 * 19 = 798
Variance of x = ((Σx^2)/19) - mean of x squared
So ((Σx^2)/19) - 42^2 = 16
So Σx^2 = (16 + 42^2) * 19 = 33820
A group of 19 people played a game. The scores, x, that the people achieved are summarised by:
Σ(x-30) = 228 and Σ(x-30)^2 = 3040
Another student played the game. Her score was 32. Find the new mean and standard deviation of all 20 scores
New Σx = 798 + 32 = 830
So new mean of x = (830/20) = 41.5
New Σx^2 = 33820 + 32^2 = 34844
So new variance of x = (34844/20) - 41.5^2 = 19.95
New standard deviation of x = 4.47
