Question

four distributions, labeled (a), (b), (c), and (d) are represented below by their histograms. each distribution is made of 9 measurements. without performing any calculations, order their respective means (mu_a), (mu_b), (mu_c), and (mu_d).
enter the four subscripts appropriately below.

Explanation:

Step1: Analyze Histogram (c)

In histogram (c), the leftmost bar (smallest value) is at 1, and there's a bar at 4. The lower values (left side) have some weight, but let's compare with others. The mean here will be pulled left a bit, but let's check others.

Step2: Analyze Histogram (a)

Histogram (a) has leftmost bar at 2, next at 4. Compared to (c), (a) starts at 2 instead of 1, so its mean should be higher than (c) since the smallest value is larger.

Step3: Analyze Histogram (b)

Histogram (b) has leftmost bar at 2, next at 4, then 7, 10, 12. Wait, no, looking at the x - axis: (b) has bars at 2, 4, 7, 10, 12. Wait, (a) has bars at 2, 4, 7, 9, 12? Wait, no, let's re - examine. The key is the position of the bars (the x - values) and their heights (frequency). For the mean, we can think of the "center of mass" of the histogram. The more weight (higher bars) on the right, the higher the mean; more weight on the left, lower the mean.

Histogram (c): Smallest x - value is 1, then 4, 7, 9, 12. Histogram (a): Smallest x - value is 2, then 4, 7, 9, 12. So (c) has a smaller x - value (1) contributing, so $\mu_c<\mu_a$? Wait, no, wait (d): Let's look at (d). Histogram (d) has bars at 4, 7, 9, 12. Wait, (d) has a bar at 4 (height 3), 7 (height 1), 9 (height 3), 12 (height 2). (b) has bars at 2 (height 1), 4 (height 2), 7 (height 1), 10 (height 3), 12 (height 2). (a) has bars at 2 (height 1), 4 (height 2), 7 (height 1), 9 (height 3), 10 (height 1), 12 (height 1). (c) has bars at 1 (height 1), 4 (height 2), 7 (height 1), 9 (height 3), 10 (height 1), 12 (height 1).

So, for (c): the smallest x is 1, so it has a lower value contributing. For (a): smallest x is 2, so higher than (c). For (b): the peak is at 10 (height 3), and (d): peak at 4 and 9 (height 3 each), but (d) has no bar at 2, while (b) has a bar at 2. Wait, no, let's think about the overall "center".

Histogram (c): Values are 1 (1 time), 4 (2 times), 7 (1 time), 9 (3 times), 10 (1 time), 12 (1 time). Total 1 + 2+1 + 3+1 + 1=9.

Histogram (a): Values are 2 (1 time), 4 (2 times), 7 (1 time), 9 (3 times), 10 (1 time), 12 (1 time). Total 1+2 + 1+3+1+1 = 9.

Histogram (b): Values are 2 (1 time), 4 (2 times), 7 (1 time), 10 (3 times), 12 (2 times). Total 1+2 + 1+3+2=9.

Histogram (d): Values are 4 (3 times), 7 (1 time), 9 (3 times), 12 (2 times). Total 3+1+3+2 = 9.

Now, let's calculate the "weighted" position (even without exact calculation, we can compare).

For (c): The sum of (x * frequency) will have a 1*1 term.

For (a): Sum has 2*1 term (instead of 1*1 in (c)), so (a)'s sum is larger than (c)'s, so $\mu_c<\mu_a$.

For (b): The peak is at 10 (frequency 3) and 12 (frequency 2), while (a) has peak at 9 (frequency 3). So (b) has more weight on the right (10 and 12) than (a) (9 and 10, 12). So $\mu_a<\mu_b$.

For (d): Let's see, (d) has values at 4 (3), 7 (1), 9 (3), 12 (2). (b) has values at 2 (1), 4 (2), 7 (1), 10 (3), 12 (2). The 2 in (b) is a small value, but (d) has 4 (3 times) which is a bit left, but (b) has 2 (1 time) and 10 (3 times) vs (d)'s 9 (3 times) and 12 (2 times). Wait, no, let's think of the mean as the average. Let's list the x - values with their frequencies:

- (c): x = 1 (1), 4 (2), 7 (1), 9 (3), 10 (1), 12 (1). Sum = 1*1+4*2 + 7*1+9*3+10*1+12*1=1 + 8+7 + 27+10+12 = 65. Mean = 65/9≈7.22
- (a): x = 2 (1), 4 (2), 7 (1), 9 (3), 10 (1), 12 (1). Sum = 2*1+4*2 + 7*1+9*3+10*1+12*1=2 + 8+7 + 27+10+12 = 66. Mean = 66/9≈7.33
- (b): x = 2 (1), 4 (2), 7 (1), 10 (3), 12 (2). Sum = 2*1+4*2 + 7*1+10*3+12*2=2 + 8+7 + 30+24 = 71. Mean = 71/9≈7.89
- (d): x = 4 (3), 7 (1), 9 (3), 12 (2). Sum = 4*3+7*1+9*3+12*2=12 + 7+27+24 = 70. Mean = 70/9≈7.78

Wait, but this contradicts? Wait, no, maybe my frequency counting is wrong. Let's look at the histograms again.

Looking at the histograms:

(a): Bars at x = 2 (height 1), x = 4 (height 2), x = 7 (height 1), x = 9 (height 3), x = 10 (height 1), x = 12 (height 1). Total height:1 + 2+1 + 3+1 + 1=9. Correct.

(b): Bars at x = 2 (height 1), x = 4 (height 2), x = 7 (height 1), x = 10 (height 3), x = 12 (height 2). Total height:1 + 2+1 + 3+2=9. Correct.

(c): Bars at x = 1 (height 1), x = 4 (height 2), x = 7 (height 1), x = 9 (height 3), x = 10 (height 1), x = 12 (height 1). Total height:1 + 2+1 + 3+1 + 1=9. Correct.

(d): Bars at x = 4 (height 3), x = 7 (height 1), x = 9 (height 3), x = 12 (height 2). Total height:3 + 1+3 + 2=9. Correct.

Now, calculating the sums:

(c): 1*1 + 4*2+7*1 + 9*3+10*1+12*1=1 + 8+7 + 27+10+12 = 65. $\mu_c=\frac{65}{9}\approx7.22$

(a): 2*1+4*2 + 7*1+9*3+10*1+12*1=2 + 8+7 + 27+10+12 = 66. $\mu_a=\frac{66}{9}\approx7.33$

(d): 4*3+7*1+9*3+12*2=12 + 7+27+24 = 70. $\mu_d=\frac{70}{9}\approx7.78$

(b): 2*1+4*2 + 7*1+10*3+12*2=2 + 8+7 + 30+24 = 71. $\mu_b=\frac{71}{9}\approx7.89$

So the order from smallest to largest mean is $\mu_c<\mu_a<\mu_d<\mu_b$? Wait, no, wait (d) sum is 70, (b) sum is 71, (a) sum 66, (c) sum 65. So 65 < 66 < 70 < 71, so $\mu_c<\mu_a<\mu_d<\mu_b$? But that seems off. Wait, maybe I misread the histograms. Let's re - examine the x - axis labels.

Wait, in (c), the first bar is at x = 1, (a) at x = 2, (b) at x = 2, (d) at x = 4.

Wait, (d) has no bar at x = 2 or 1. So (d)'s smallest x is 4, (a) and (b) have x = 2, (c) has x = 1.

So (c) has the smallest x (1), so its mean is the smallest. Then (a) has x = 2 (smaller than (d)'s x = 4), so (a) mean is next. Then (d) has x = 4, and (b) has x = 2 but also has a bar at x = 10 (higher than (d)'s x = 9). Wait, no, (d) has a bar at x = 9 (height 3) and (b) at x = 10 (height 3). (d) has a bar at x = 12 (height 2) and (b) at x = 12 (height 2). (d) has a bar at x = 4 (height 3) and (b) at x = 4 (height 2). (d) has a bar at x = 7 (height 1) and (b) at x = 7 (height 1). (b) has a bar at x = 2 (height 1) and (d) has no bar at x = 2.

So the sum for (b): 2*1 (from x = 2) + 4*2 (x = 4) + 7*1 (x = 7) + 10*3 (x = 10) + 12*2 (x = 12) = 2+8 + 7+30 + 24=71

Sum for (d):4*3 (x = 4) + 7*1 (x = 7) + 9*3 (x = 9) + 12*2 (x = 12)=12 + 7+27 + 24=70

Sum for (a):2*1 (x = 2) + 4*2 (x = 4) + 7*1 (x = 7) + 9*3 (x = 9) + 10*1 (x = 10) + 12*1 (x = 12)=2+8 + 7+27+10 + 12=66

Sum for (c):1*1 (x = 1) + 4*2 (x = 4) + 7*1 (x = 7) + 9*3 (x = 9) + 10*1 (x = 10) + 12*1 (x = 12)=1+8 + 7+27+10 + 12=65

So the order of means from smallest to largest is $\mu_c<\mu_a<\mu_d<\mu_b$? Wait, but (d) has a sum of 70, (a) 66, so 66 < 70, so $\mu_a<\mu_d$, and $\mu_d<\mu_b$ (70 < 71), and $\mu_c<\mu_a$ (65 < 66). So the order is $\mu_c<\mu_a<\mu_d<\mu_b$? But that seems to be the case from the sums.

Wait, but maybe the initial analysis was wrong. Let's confirm with the "center of mass" idea. The mean is like the balance point. A histogram with more weight on the left has a smaller mean, more on the right has a larger mean.

- (c) has weight at x = 1 (leftmost), so balance point leftmost.
- (a) has weight at x = 2 (right of (c)'s x = 1), so balance point right of (c).
- (d) has weight starting at x = 4 (right of (a)'s x = 2), so balance point right of (a).
- (b) has weight at x = 10 (right of (d)'s x = 9) and x = 12 (same as (d)), but also has weight at x = 2 (left of (d)'s x = 4), but the weight at x = 10 (height 3) is more right - shifted than (d)'s x = 9 (height 3), so (b)'s balance point is right of (d)'s.

So the order of the means is $\mu_c<\mu_a<\mu_d<\mu_b$

Answer:

$\mu_{c}<\mu_{a}<\mu_{d}<\mu_{b}$