Question

the number of bears killed in 2010 for 52 counties in pennsylvania is shown in the frequency distribution.

class limits frequency
1 – 25 18
26 – 50 12
51 – 75 7
76 – 100 6
101 – 125 2
126 – 150 3
151 – 175 0
176 – 200 2
201 – 225 0
226 – 250 2
total 52

send data to excel

part: 0 / 2

Explanation:

Since the problem is not fully stated (only the frequency distribution table is provided without a specific question like finding mean, median, mode, or something else), we can't provide a solution yet. If you want to find, for example, the mean number of bears killed, the steps would be:

Step 1: Find the midpoint of each class

For class \(1 - 25\), midpoint \(x_1=\frac{1 + 25}{2}=13\)
For class \(26 - 50\), midpoint \(x_2=\frac{26+50}{2} = 38\)
For class \(51 - 75\), midpoint \(x_3=\frac{51 + 75}{2}=63\)
For class \(76 - 100\), midpoint \(x_4=\frac{76+100}{2}=88\)
For class \(101 - 125\), midpoint \(x_5=\frac{101+125}{2}=113\)
For class \(126 - 150\), midpoint \(x_6=\frac{126+150}{2}=138\)
For class \(151 - 175\), midpoint \(x_7=\frac{151+175}{2}=163\)
For class \(176 - 200\), midpoint \(x_8=\frac{176+200}{2}=188\)
For class \(201 - 225\), midpoint \(x_9=\frac{201+225}{2}=213\)
For class \(226 - 250\), midpoint \(x_{10}=\frac{226+250}{2}=238\)

Step 2: Multiply each midpoint by its frequency

\(f_1x_1=18\times13 = 234\)
\(f_2x_2=12\times38=456\)
\(f_3x_3=7\times63 = 441\)
\(f_4x_4=6\times88=528\)
\(f_5x_5=2\times113 = 226\)
\(f_6x_6=3\times138=414\)
\(f_7x_7=0\times163 = 0\)
\(f_8x_8=2\times188=376\)
\(f_9x_9=0\times213 = 0\)
\(f_{10}x_{10}=2\times238 = 476\)

Step 3: Sum up all the \(f_ix_i\) values

\(\sum f_ix_i=234 + 456+441+528+226+414+0+376+0+476\)
\(=234+456 = 690\); \(690+441=1131\); \(1131 + 528=1659\); \(1659+226=1885\); \(1885+414=2299\); \(2299+376=2675\); \(2675 + 476=3151\)

Step 4: Divide by the total frequency (\(N = 52\))

Mean \(\bar{x}=\frac{\sum f_ix_i}{N}=\frac{3151}{52}\approx60.60\)

But since the actual question is not given, please provide the specific question (like find the mean, median, mode, or construct a histogram etc.) so that we can give a more targeted solution.

Answer:

Since the problem is not fully stated (only the frequency distribution table is provided without a specific question like finding mean, median, mode, or something else), we can't provide a solution yet. If you want to find, for example, the mean number of bears killed, the steps would be:

Step 1: Find the midpoint of each class

For class \(1 - 25\), midpoint \(x_1=\frac{1 + 25}{2}=13\)
For class \(26 - 50\), midpoint \(x_2=\frac{26+50}{2} = 38\)
For class \(51 - 75\), midpoint \(x_3=\frac{51 + 75}{2}=63\)
For class \(76 - 100\), midpoint \(x_4=\frac{76+100}{2}=88\)
For class \(101 - 125\), midpoint \(x_5=\frac{101+125}{2}=113\)
For class \(126 - 150\), midpoint \(x_6=\frac{126+150}{2}=138\)
For class \(151 - 175\), midpoint \(x_7=\frac{151+175}{2}=163\)
For class \(176 - 200\), midpoint \(x_8=\frac{176+200}{2}=188\)
For class \(201 - 225\), midpoint \(x_9=\frac{201+225}{2}=213\)
For class \(226 - 250\), midpoint \(x_{10}=\frac{226+250}{2}=238\)

Step 2: Multiply each midpoint by its frequency

\(f_1x_1=18\times13 = 234\)
\(f_2x_2=12\times38=456\)
\(f_3x_3=7\times63 = 441\)
\(f_4x_4=6\times88=528\)
\(f_5x_5=2\times113 = 226\)
\(f_6x_6=3\times138=414\)
\(f_7x_7=0\times163 = 0\)
\(f_8x_8=2\times188=376\)
\(f_9x_9=0\times213 = 0\)
\(f_{10}x_{10}=2\times238 = 476\)

Step 3: Sum up all the \(f_ix_i\) values

\(\sum f_ix_i=234 + 456+441+528+226+414+0+376+0+476\)
\(=234+456 = 690\); \(690+441=1131\); \(1131 + 528=1659\); \(1659+226=1885\); \(1885+414=2299\); \(2299+376=2675\); \(2675 + 476=3151\)

Step 4: Divide by the total frequency (\(N = 52\))

Mean \(\bar{x}=\frac{\sum f_ix_i}{N}=\frac{3151}{52}\approx60.60\)

But since the actual question is not given, please provide the specific question (like find the mean, median, mode, or construct a histogram etc.) so that we can give a more targeted solution.