Question

final exam - practice

1. frequency distribution table

a survey was given to a group of college students, asking them how many hours per day, on average, they spent on their electronic devices. this table details their responses. find the missing values. round the relative frequency and cumulative relative frequency values to the thousandths place.
table: screen time (hours) with columns f, rf, cf, rcf. rows: 0 - 1 (f=15, rf=0.085, cf=b, rcf=0.085); 1 - 2 (f=19, rf=0.107, cf=34, rcf=0.192); 2 - 3 (f=27, rf=0.153, cf=61, rcf=0.345); 3 - 4 (f=35, rf=0.198, cf=96, rcf=0.542); 4 - 5 (f=20, rf=0.113, cf=116, rcf=d); 5 - 6 (f=21, rf=0.119, cf=137, rcf=0.774); 6 - 7 (f=16, rf=0.090, cf=c, rcf=0.864); 7 - 8 (f=a, rf=0.062, cf=164, rcf=0.927); 8+ (f=13, rf=0.073, cf=177, rcf=1.000)

1. a. = ____ 2) b. = __ 3) c. = __ 4) d. = ____
2. how many students logged at least 6 hours of screen time? ______
3. what percentage of students logged no more than 4 hours of screen time? ______

a survey was given to a group of college students, asking them how many hours per day, on average, they spent streaming shows and/or films. this table details their responses. find the missing values. round the relative frequency and cumulative relative frequency values to the thousandths place.
table: streaming time (hours) with columns f, rf, cf, rcf. rows: 0 - 1 (f=15, rf=0.073, cf=15, rcf=0.073); 1 - 2 (f=21, rf=a, cf=36, rcf=0.176); 2 - 3 (f=30, rf=b, cf=66, rcf=0.323); 3 - 4 (f=32, rf=0.156, cf=98, rcf=0.478); 4 - 5 (f=31, rf=0.151, cf=129, rcf=d); 5 - 6 (f=43, rf=0.209, cf=172, rcf=0.839); 6 - 7 (f=11, rf=0.054, cf=183, rcf=0.893); 7 - 8 (f=9, rf=0.044, cf=192, rcf=0.937); 8+ (f=13, rf=0.063, cf=c, rcf=1.000)

1. a. = ____ 8) b. = __ 9) c. = __ 10) d. = ____
2. how many students streamed no more than 7 hours? ______
3. what percentage of students streamed at least 3 hours? ______

Explanation:

Step 1: Find the total number of students (for the first table)

The last cumulative frequency (cf) is 177, so the total number of students \( N = 177 \).

Step 1 (a): Find \( a \) (frequency for 7 - 8 hours)

We know that relative frequency \( rf=\frac{f}{N} \). For the 7 - 8 hours category, \( rf = 0.062 \) and \( N = 177 \). So \( f=a=\text{rf}\times N=0.062\times177 \).
\( 0.062\times177 = 10.974\approx11 \) (wait, maybe I made a mistake. Wait, the cumulative frequency before 7 - 8 is 154 (from 6 - 7: cf = c, then 7 - 8: cf = 164? Wait, the table says 7 - 8 has cf = 164? Wait, the first table:

Screen Time (hours) | f | rf | cf | rcf
--- | --- | --- | --- | ---
0 - 1 | 15 | 0.085 | b | 0.085
1 - 2 | 19 | 0.107 | 34 | 0.192
2 - 3 | 27 | 0.153 | 61 | 0.345
3 - 4 | 35 | 0.198 | 96 | 0.542
4 - 5 | 20 | 0.113 | 116 | d
5 - 6 | 21 | 0.119 | 137 | 0.774
6 - 7 | 16 | 0.090 | c | 0.864
7 - 8 | a | 0.062 | 164 | 0.927
8 + | 13 | 0.073 | 177 | 1.000

So cumulative frequency for 6 - 7: c. Then 7 - 8: cf = c + a = 164. And 8 +: cf = 164 + 13 = 177. So c = 164 - a. Also, rcf for 6 - 7 is 0.864, and for 7 - 8 is 0.927, so the relative frequency for 7 - 8 is 0.927 - 0.864 = 0.063? Wait, the table says rf for 7 - 8 is 0.062. Maybe the total N is 177. Let's check the sum of frequencies: 15 + 19 + 27 + 35 + 20 + 21 + 16 + a + 13 = 177.

Sum of known f: 15 + 19 = 34; 34 + 27 = 61; 61 + 35 = 96; 96 + 20 = 116; 116 + 21 = 137; 137 + 16 = 153; 153 + a + 13 = 177. So 153 + 13 + a = 177 → 166 + a = 177 → a = 11.

Step 1 (b): Find \( b \) (cf for 0 - 1)

Cumulative frequency (cf) for 0 - 1 is the frequency of 0 - 1, since it's the first category. So \( b = 15 \).

Step 1 (c): Find \( c \) (cf for 6 - 7)

cf for 6 - 7 is cf of 5 - 6 + f of 6 - 7. cf of 5 - 6 is 137, f of 6 - 7 is 16. So \( c = 137 + 16 = 153 \). Wait, but the table says cf for 7 - 8 is 164. 153 + 11 (a) = 164, which matches.

Step 1 (d): Find \( d \) (rcf for 4 - 5)

rcf is cumulative relative frequency, so it's the sum of relative frequencies up to 4 - 5. The relative frequencies up to 3 - 4 is 0.542, and for 4 - 5 is 0.113. So \( d = 0.542 + 0.113 = 0.655 \).

Step 2: Question 5: How many students logged at least 6 hours of screen time?

"At least 6 hours" means 6 - 7, 7 - 8, 8 +. So sum their frequencies: 16 (6 - 7) + 11 (7 - 8) + 13 (8 +) = 16 + 11 + 13 = 40.

Step 3: Question 6: Percentage of students with no more than 4 hours of screen time?

"No more than 4 hours" means 0 - 1, 1 - 2, 2 - 3, 3 - 4. The cumulative relative frequency for 3 - 4 is 0.542. So percentage is \( 0.542\times100 = 54.2\% \).

Now for the second table (streaming time):

Streaming Time (hours) | f | rf | cf | rcf
--- | --- | --- | --- | ---
0 - 1 | 15 | 0.073 | 15 | 0.073
1 - 2 | 21 | a | 36 | 0.176
2 - 3 | 30 | b | 66 | 0.323
3 - 4 | 32 | 0.156 | 98 | 0.478
4 - 5 | 31 | 0.151 | 129 | d
5 - 6 | 43 | 0.209 | 172 | 0.839
6 - 7 | 11 | 0.054 | 183 | 0.893
7 - 8 | 9 | 0.044 | 192 | 0.937
8 + | 13 | 0.063 | c | 1.000

Step 2 (a): Find \( a \) (rf for 1 - 2)

rf = f / N. First, find N. The last cf is c, and 8 + has f = 13, rf = 0.063, so \( N=\frac{13}{0.063}\approx206.35 \). Wait, but cf for 7 - 8 is 192, 8 +: cf = 192 + 13 = 205? Wait, the table says 8 + has cf = c and rcf = 1.000. Let's check the sum of frequencies:

0 - 1:15, 1 - 2:21, 2 - 3:30, 3 - 4:32, 4 - 5:31, 5 - 6:43, 6 - 7:11, 7 - 8:9, 8 +:13.

Sum: 15 + 21 = 36; 36 + 30 = 66; 66 + 32 = 98; 98 + 31 = 129; 129 + 43 = 172; 172 + 11 = 183; 183 + 9 = 192; 192 + 13 = 205. So N = 205.

So for 1 - 2: \( a=\frac{21}{205}\approx0.102 \) (since rcf for 1 - 2 is 0.176, and rcf for 0 - 1 is 0.073, so 0.176 - 0.073 = 0.103? Wait, 21/205 = 0.1024... ≈ 0.102.

Step 2 (b): Find \( b \) (rf for 2 - 3)

rf = f / N = 30 / 205 ≈ 0.146. Or rcf for 2 - 3 is 0.323, rcf for 1 - 2 is 0.176, so 0.323 - 0.176 = 0.147. 30/205 = 0.1463... ≈ 0.146.

Step 2 (c): Find \( c \) (cf for 8 +)

cf for 8 + is the sum of all frequencies, which is 205. So \( c = 205 \).

Step 2 (d): Find \( d \) (rcf for 4 - 5)

rcf is cumulative relative frequency, so sum of relative frequencies up to 4 - 5. Relative frequencies up to 3 - 4 is 0.478, and for 4 - 5 is 0.151. So \( d = 0.478 + 0.151 = 0.629 \).

Step 3: Question 11: How many students streamed no more than 7 hours?

"No more than 7 hours" means up to 7 - 8. The cf for 7 - 8 is 192. So 192 students.

Step 4: Question 12: Percentage of students who streamed at least 3 hours?

"At least 3 hours" means 3 - 4, 4 - 5, 5 - 6, 6 - 7, 7 - 8, 8 +. The cumulative relative frequency for 2 - 3 is 0.323, so the relative frequency for at least 3 hours is \( 1 - 0.323 = 0.677 \). Percentage is \( 0.677\times100 = 67.7\% \).

Wait, let's re - do the first table's a:

First table, 7 - 8: rf = 0.062, N = 177 (since 8 + has f = 13, rf = 0.073, 13/0.073≈178.08, but cf is 177, so N = 177. Then a = rf×N = 0.062×177 = 10.974≈11. Which matches 153 (c) + 11 = 164, and 164 + 13 = 177.

Now, let's summarize the answers for the first set (1 - 6):

1) a = 11

2) b = 15 (cf for 0 - 1 is f = 15)

3) c = 153 (137 + 16)

4) d = 0.655 (0.542 + 0.113)

5) At least 6 hours: 16 + 11 + 13 = 40

6) No more than 4 hours: 0.542×100 = 54.2%

For the second set (7 - 12):

7) a = 21/205≈0.102 (or 0.103 if using rcf: 0.176 - 0.073 = 0.103)

8) b = 30/205≈0.146 (or 0.323 - 0.176 = 0.147)

9) c = 205

10) d = 0.629 (0.478 + 0.151)

11) No more than 7 hours: 192

12) At least 3 hours: (1 - 0.323)×100 = 67.7%

But let's check the first table's a again. The rf for 7 - 8 is 0.062, N = 177, so a = 0.062×177 = 10.974, which we can round to 11.

For the first table's b: cf for 0 - 1 is the frequency of 0 - 1, which is 15. So b = 15.

For the first table's c: cf for 6 - 7 is cf of 5 - 6 (137) + f of 6 - 7 (16) = 153. So c = 153.

For the first table's d: rcf for 4 - 5 is rcf of 3 - 4 (0.542) + rf of 4 - 5 (0.113) = 0.655. So d = 0.655.

Question 5: At least 6 hours: 6 - 7 (16) + 7 - 8 (11) + 8 + (13) = 16 + 11 + 13 = 40.

Question 6: No more than 4 hours: cumulative relative frequency up to 3 - 4 is 0.542, so 54.2%.

For the second table:

N = 15 + 21 + 30 + 32 + 31 + 43 + 11 + 9 + 13 = 205.

7) a: rf for 1 - 2 is f/N = 21/205≈0.102 (or 0.103 if we use rcf: 0.176 - 0.073 = 0.103). Let's check 21/205 = 0.1024, which rounds to 0.102.

8) b: rf for 2 - 3 is 30/205≈0.146 (or 0.323 - 0.176 = 0.147). 30/205 = 0.1463, rounds to 0.146.

9) c: cf for 8 + is the total number of students, which is 205. So c = 205.

10) d: rcf for 4 - 5 is rcf of 3 - 4 (0.478) + rf of 4 - 5 (0.151) = 0.629.

11) No more than 7 hours: cf for 7 - 8 is 192, so 192 students.

12) At least 3 hours: relative frequency is 1 - rcf of 2 - 3 (0.323) = 0.677, so percentage is 67.7%.

Answer:

1. \( a = \boldsymbol{11} \)

1. \( b = \boldsymbol{15} \)

1. \( c = \boldsymbol{153} \)

1. \( d = \boldsymbol{0.655} \)

1. \(\boldsymbol{40}\)

1. \(\boldsymbol{54.2\%}\)

1. \( a = \boldsymbol{0.102} \) (or 0.103)

1. \( b = \boldsymbol{0.146} \) (or 0.147)

1. \( c = \boldsymbol{205} \)

1. \( d = \boldsymbol{0.629} \)

1. \(\boldsymbol{192}\)

1. \(\boldsymbol{67.7\%}\)