Question

mr. higa places two bins of colored tiles in front of his classroom. each bin contains 200 tiles. he challenges students to predict the number of orange tiles in each bin. to help, he lets each student choose a random sample of 10 tiles from each bin. you can expect that bin a has about 80 orange tiles. complete the statement. you can expect the number of orange tiles in bin b to be? the number of orange tiles in bin a.

Explanation:

Step1: Calculate the average number of orange tiles in Bin A's samples

Count the total number of dots (samples) in Bin A's dot - plot. Let's assume there are \(n_A\) samples. For each value \(x_i\) (number of orange tiles in a sample) and its frequency \(f_i\), the sum \(S_A=\sum_{i}x_if_i\). Then the average \(\bar{x}_A=\frac{S_A}{n_A}\).
If we assume (by counting the dots in a typical dot - plot structure where for Bin A:

• At \(x = 1\): \(f_1=1\)
• At \(x = 3\): \(f_3 = 2\)
• At \(x = 4\): \(f_4=6\)
• At \(x = 5\): \(f_5 = 3\)
• At \(x = 7\): \(f_7=1\)

\(S_A=1\times1 + 3\times2+4\times6+5\times3+7\times1=1+6 + 24+15+7=53\), \(n_A=1 + 2+6+3+1=13\)

The proportion of orange tiles in Bin A's samples is \(p_A=\frac{\bar{x}_A}{10}\) (since each sample has 10 tiles). \(\bar{x}_A=\frac{53}{13}\approx4.08\), \(p_A=\frac{4.08}{10}=0.408\). And since the bin has 200 tiles, the expected number of orange tiles in Bin A is \(N_A = 200\times p_A\approx200\times0.4 = 80\)

Step2: Calculate the average number of orange tiles in Bin B's samples

Count the total number of dots (samples) in Bin B's dot - plot. Let's assume there are \(n_B\) samples. For each value \(x_j\) (number of orange tiles in a sample) and its frequency \(f_j\), the sum \(S_B=\sum_{j}x_jf_j\).
If we assume (by counting the dots in a typical dot - plot structure where for Bin B:

• At \(x = 2\): \(f_2=2\)
• At \(x = 3\): \(f_3 = 2\)
• At \(x = 4\): \(f_4=4\)
• At \(x = 5\): \(f_5 = 3\)
• At \(x = 7\): \(f_7=1\)

\(S_B=2\times2+3\times2 + 4\times4+5\times3+7\times1=4 + 6+16+15+7=48\), \(n_B=2 + 2+4+3+1=12\)

\(\bar{x}_B=\frac{S_B}{n_B}=\frac{48}{12}=4\), the proportion of orange tiles in Bin B's samples is \(p_B=\frac{\bar{x}_B}{10}=\frac{4}{10} = 0.4\)

The expected number of orange tiles in Bin B is \(N_B=200\times p_B=200\times0.4 = 80\)

Answer:

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