Question

the number of hospital beds in a sample of 20 hospitals is shown below. construct a frequency distribution and a frequency histogram for the data set using 5 classes. describe the shape of the histogram as symmetric, uniform, negatively skewed, positively skewed, or none of these. 171 160 131 129 176 158 165 216 148 138 195 212 153 254 257 245 309 134 204 180. construct a frequency distribution for the data set using 5 classes. (type whole numbers.) class frequency 129–165 □ 166–202 □ 203–239 □ 240–276 □ 277–313 □

Explanation:

Step1: Identify data points

The data points are: 171, 160, 131, 129, 176, 158, 165, 216, 148, 138, 195, 212, 153, 254, 257, 245, 309, 134, 204, 180.

Step2: Count for 129–165

Check which numbers are between 129 (inclusive) and 165 (inclusive). The numbers are: 131, 129, 160, 158, 165, 148, 138, 153, 134. Let's count: 129 (1), 131 (2), 134 (3), 138 (4), 148 (5), 153 (6), 158 (7), 160 (8), 165 (9). So frequency is 9.

Step3: Count for 166–202

Numbers between 166 and 202 (inclusive): 171, 176, 195, 180. Wait, 171 (1), 176 (2), 180 (3), 195 (4). Wait, did I miss? Wait 166–202: 171 (yes), 176 (yes), 195 (yes), 180 (yes), and what about 204? No, 204 is above 202. Wait, original data: 171, 160 (no, 160 is in first class), 176, 195, 180. Wait, also 166–202: 171, 176, 180, 195. Wait, is there another? Wait 165 is in first class, 171 is 171 ≥166, ≤202. 176 same, 180, 195. Wait, 204 is 204>202, so no. Wait, how many? 171,176,180,195. Wait, that's 4? Wait no, wait the data points: 171, 160, 131, 129, 176, 158, 165, 216, 148, 138, 195, 212, 153, 254, 257, 245, 309, 134, 204, 180. Let's list all in 166–202: 171 (1), 176 (2), 180 (3), 195 (4). Wait, is 204 in 203–239. So yes, 4 here? Wait no, wait 166–202: 171,176,180,195. Wait, 204 is 204, which is 203–239. So frequency 4? Wait no, wait maybe I made a mistake. Wait 166–202: numbers from 166 to 202. Let's check each data point:

171: 171 is between 166-202: yes.

160: no (160 <166).

131: no.

129: no.

176: yes.

158: no.

165: no (165 <166).

216: no (216>202).

148: no.

138: no.

195: yes.

212: no.

153: no.

254: no.

257: no.

245: no.

309: no.

134: no.

204: no (204>202).

180: yes.

So that's 171,176,195,180: 4 numbers? Wait, but 171,176,180,195: that's 4. Wait, but maybe I missed? Wait 166–202: 166 ≤ x ≤202. So 171,176,180,195: 4. So frequency 4?

Wait no, wait 171, 176, 180, 195: that's 4. Wait, but let's check again. Wait 166–202: 171 (yes), 176 (yes), 180 (yes), 195 (yes). So 4.

Step4: Count for 203–239

Numbers between 203 and 239 (inclusive): 216, 212, 204. Let's check:

216: yes (203 ≤216 ≤239).

212: yes.

204: yes.

Any others? 254: no (254>239). 216,212,204: that's 3. Wait, 216,212,204: 3 numbers. So frequency 3.

Step5: Count for 240–276

Numbers between 240 and 276 (inclusive): 254, 257, 245. Let's check:

245: yes (240 ≤245 ≤276).

254: yes.

257: yes.

Any others? 309: no (309>276). So 245,254,257: 3 numbers. So frequency 3.

Step6: Count for 277–313

Numbers between 277 and 313 (inclusive): 309. Let's check: 309 is between 277 and 313. Any others? 257: no (257<277). So only 309. So frequency 1.

Wait, but let's sum the frequencies: 9 + 4 + 3 + 3 + 1 = 20, which matches the number of data points (20 hospitals). Good.

Wait, but wait step 3: 166–202. Wait, 171,176,180,195: that's 4. But wait, 166–202: 166 to 202. Let's check 195: 195 is ≤202. Yes. 180: yes. 176: yes. 171: yes. So 4. Correct.

Step 4: 203–239: 204,212,216: 3. Correct.

Step 5: 240–276: 245,254,257: 3. Correct.

Step 6: 277–313: 309: 1. Correct.

Step 2: 129–165: let's list all data points in this range:

129 (yes), 131 (yes), 134 (yes), 138 (yes), 148 (yes), 153 (yes), 158 (yes), 160 (yes), 165 (yes). Wait, that's 9 numbers. Let's count:

1. 129

1. 131

1. 134

1. 138

1. 148

1. 153

1. 158

1. 160

1. 165

Yes, 9. Correct.

So the frequencies are:

129–165: 9

166–202: 4

203–239: 3

240–276: 3

277–313: 1

Answer:

Class	Frequency
166–202	4
203–239	3
240–276	3
277–313	1

For the histogram shape: The frequencies start high (9), then decrease (4, 3, 3), then a low (1) at the higher end. So the tail is on the right (higher values), so it's positively skewed. But the question first asks for the frequency distribution, which we provided.