Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 56.6 degrees. Low Temperature (◦F) 40−44 45−49 50−54 55−59 60−64 Frequency 2 7 9 5 2 The mean of the frequency distribution is nothing degrees.

1. The mean of the data summarized in the given frequency distribution is 51.81 degrees

2. Comparing the computed mean (51.81 degrees) to the actual mean(56.6) degrees, the computed mean is lesser than the actual mean.

Step-by-step explanation:

1) We are given a Frequency distribution.

Mean of a frequency distribution = Fx/F

Where

Fx = Frequency × midpoint

Low Temperature (◦F) Frequency

40−44 2

45−49 7

50−54 9

55−59 6

60−64 2

F =Total Frequency

= 2 + 7 + 9 + 6 + 2

= 26

a)For low temperature 40 - 44

Frequency = 2

Midpoint (x) = 40 +44/2 = 42

Fx =42 × 2 = 84

b) For low temperature 45 - 49

Frequency = 2

Midpoint (x) = 45+49/2 = 47

Fx =47 × 7 = 329

For low temperature 50 - 54

Frequency = 2

Midpoint (x) = 50 + 54/2 = 52

Fx =52 × 9 = 468

For low temperature 55 - 59

Frequency = 6

Midpoint (x) = 55 + 59/2 = 57

Fx =57 × 6 = 342

For low temperature 60 - 64

Frequency = 2

Midpoint (x) = 60 +64/2 = 62

Fx =62 × 2 = 124

Total Fx =84 + 329 + 468 + 342 + 124

= 1347

Mean =Total Fx/ Total Frequency

= 1347/26

= 51.807692308

Approximately = 51.81 degrees

Therefore, mean of the data summarized in the given frequency distribution is 51.81 degrees

2. Comparing the computed mean (51.81 degrees) to the actual mean(56.6) degrees, the computed mean is lesser than the actual mean.

26

step-by-step explanation:

since proposition states that a number is greater than 25, if is true then the number must actually be greater than 25.

this implies that all possible options, except 24, can be accepted