Question

as of 2011, the following are the ages at which a countrys presidents were inaugurated.
53, 60, 52, 62, 60, 55, 51, 47, 48, 49, 62, 50, 63, 69, 68, 62, 69, 67,
44, 46, 64, 48, 47, 59, 55, 60, 68, 60, 61, 61, 57, 47, 56, 49, 46, 48,
61, 64, 44, 51, 58, 59, 55, 42
construct a grouped frequency distribution for the data. use 41 - 45 for the first class and use the same width for each subsequent class.
type the frequency value for each class.
age
41 - 45
46 - 50
frequency
3

Explanation:

Step1: Determine class width

The first class is 41 - 45, so the class width is \( 45 - 41 + 1 = 5 \)? Wait, no, actually for grouped frequency, the class width is the difference between the upper and lower bounds. Wait, 41 - 45, so the width is \( 45 - 41 = 4 \)? Wait, no, maybe the class width is 5 (since 41 to 45 is 5 numbers: 41,42,43,44,45). Then the next class is 46 - 50 (46,47,48,49,50), so class width is 5.

Step2: Count frequencies for 46 - 50

Now, let's list all the ages and count how many are between 46 and 50 (inclusive). The ages are: 53, 60, 52, 62, 60, 55, 51, 47, 48, 49, 62, 60, 69, 68, 62, 69, 67, 44, 46, 64, 48, 47, 59, 55, 60, 68, 60, 61, 61, 57, 47, 56, 49, 46, 48, 61, 64, 44, 51, 58, 59, 55, 42. Wait, maybe I missed some. Wait, let's list all the ages:

Wait the given data: 53, 60, 52, 62, 60, 55, 51, 47, 48, 49, 62, 60, 69, 68, 62, 69, 67, 44, 46, 64, 48, 47, 59, 55, 60, 68, 60, 61, 61, 57, 47, 56, 49, 46, 48, 61, 64, 44, 51, 58, 59, 55, 42. Wait, let's count the numbers between 46 and 50 (46,47,48,49,50). Let's check each number:

46: appears (46, 46) → 2 times? Wait 46: first 46, then another 46? Wait the data: 44, 46, 64, 48, 47, 59, 55, 60, 68, 60, 61, 61, 57, 47, 56, 49, 46, 48, 61, 64, 44, 51, 58, 59, 55, 42. Wait maybe I made a mistake. Wait the original data:

Wait the user's data: "53, 60, 52, 62, 60, 55, 51, 47, 48, 49, 62, 60, 69, 68, 62, 69, 67, 44, 46, 64, 48, 47, 59, 55, 60, 68, 60, 61, 61, 57, 47, 56, 49, 46, 48, 61, 64, 44, 51, 58, 59, 55, 42" Wait, let's list all the ages and check for 46 - 50:

46: 46 (from "44, 46, 64..."), 46 (from "49, 46, 48...") → 2 times? Wait no, let's count each age:

42: 1

44: 44, 44, 44 → 3

46: 46, 46 → 2? Wait no, "44, 46, 64, 48, 47, 59, 55, 60, 68, 60, 61, 61, 57, 47, 56, 49, 46, 48, 61, 64, 44, 51, 58, 59, 55, 42" Wait, here: 46 (first), then 46 (later). So 2? Wait no, maybe I missed. Wait the original data:

Wait the initial data: "53, 60, 52, 62, 60, 55, 51, 47, 48, 49, 62, 60, 69, 68, 62, 69, 67, 44, 46, 64, 48, 47, 59, 55, 60, 68, 60, 61, 61, 57, 47, 56, 49, 46, 48, 61, 64, 44, 51, 58, 59, 55, 42" Wait, let's list all the ages:

53, 60, 52, 62, 60, 55, 51, 47, 48, 49, 62, 60, 69, 68, 62, 69, 67, 44, 46, 64, 48, 47, 59, 55, 60, 68, 60, 61, 61, 57, 47, 56, 49, 46, 48, 61, 64, 44, 51, 58, 59, 55, 42. Wait, now let's count 46 - 50:

46: 46, 46 → 2

47: 47, 47, 47 → 3

48: 48, 48, 48 → 3

49: 49, 49 → 2

50: Is there a 50? Looking at the data, do we have 50? The ages: 53, 60, 52, 62, 60, 55, 51, 47, 48, 49, 62, 60, 69, 68, 62, 69, 67, 44, 46, 64, 48, 47, 59, 55, 60, 68, 60, 61, 61, 57, 47, 56, 49, 46, 48, 61, 64, 44, 51, 58, 59, 55, 42. No 50. So 46 (2) + 47 (3) + 48 (3) + 49 (2) = 2 + 3 + 3 + 2 = 10? Wait that can't be. Wait maybe I made a mistake. Wait let's list all the ages and check:

Wait the data points:

1. 53

2. 60

3. 52

4. 62

5. 60

6. 55

7. 51

8. 47

9. 48

10. 49

11. 62

12. 60

13. 69

14. 68

15. 62

16. 69

17. 67

18. 44

19. 46

20. 64

21. 48

22. 47

23. 59

24. 55

25. 60

26. 68

27. 60

28. 61

29. 61

30. 57

31. 47

32. 56

33. 49

34. 46

35. 48

36. 61

37. 64

38. 44

39. 51

40. 58

41. 59

42. 55

43. 42

Now, let's check which are between 46 and 50 (inclusive):

- 46: data points 19, 34 → 2

- 47: data points 8, 22, 31 → 3

- 48: data points 9, 21, 35 → 3

- 49: data points 10, 33 → 2

- 50: none → 0

Total: 2 + 3 + 3 + 2 + 0 = 10. Wait, but let's check again. Wait data point 10: 49, data point 33: 49 → 2. Data point 9: 48, 21: 48, 35: 48 → 3. Data point 8: 47, 22: 47, 31: 47 → 3. Data point 19: 46, 34: 46 → 2. So total 2 + 3 + 3 + 2 = 10. So the frequency for 46 - 50 is 10.

Wait but the first class 41 - 45 has frequency 3 (ages 41 - 45: 42, 44, 44, 44? Wait 41 - 45: 42, 44, 44, 44? Wait data points: 42 (43), 44 (18), 44 (38), 44 (wait, data point 18: 44, 38: 44, any others? Wait data point 43: 42. So 41 - 45: 42, 44, 44, 44? Wait 42 is 42 (in 41 - 45), 44 is 44 (in 41 - 45). So data points 18: 44, 38: 44, 43: 42. Wait, maybe the first class 41 - 45 has ages 41,42,43,44,45. So 42, 44, 44, 44? Wait data: 42 (1), 44 (18), 44 (38), 44 (is there another 44? Wait data point 18: 44, 38: 44, any other 44? Wait data point 18: 44, 38: 44, and maybe another? Wait the data: "44, 46, 64, 48, 47, 59, 55, 60, 68, 60, 61, 61, 57, 47, 56, 49, 46, 48, 61, 64, 44, 51, 58, 59, 55, 42" Wait, here: 44 (18), 44 (38), 42 (43). So 42, 44, 44, 44? Wait 42, 44, 44, 44: that's 4 data points? But the problem says the first class 41 - 45 has frequency 3. Maybe I miscounted. Wait the problem's table shows for 41 - 45, frequency is 3. So maybe the ages in 41 - 45 are 42, 44, 44 (three data points: 42, 44, 44). So then 46 - 50: let's recount.

Wait maybe the class width is 5, so 41 - 45 (width 5: 41-45), 46 - 50 (46-50), 51 - 55, etc. Wait, let's use the class width as 5 (since 45 - 41 = 4, but maybe inclusive, so 41-45 is 5 numbers: 41,42,43,44,45; 46-50: 46,47,48,49,50; etc.

So 41 - 45: ages 41,42,43,44,45. The data has 42, 44, 44, 44? Wait no, the problem's table says frequency 3 for 41 - 45, so we can take that as given. Then for 46 - 50, we need to count how many ages are between 46 and 50 (inclusive). Let's list all the ages again, carefully:

The given ages (let's list all of them):

53, 60, 52, 62, 60, 55, 51, 47, 48, 49, 62, 60, 69, 68, 62, 69, 67, 44, 46, 64, 48, 47, 59, 55, 60, 68, 60, 61, 61, 57, 47, 56, 49, 46, 48, 61, 64, 44, 51, 58, 59, 55, 42.

Now, let's list each age and check if it's in 46 - 50:

- 46: appears at positions 19 and 34 → 2 times

- 47: appears at positions 8, 22, 31 → 3 times

- 48: appears at positions 9, 21, 35 → 3 times

- 49: appears at positions 10, 33 → 2 times

- 50: does not appear → 0 times

Now, sum these: 2 + 3 + 3 + 2 + 0 = 10. So the frequency for 46 - 50 is 10.

Answer:

10