Answer:
Neighborhood Q appears to have a bigger family size
Step-by-step explanation:
Mean = the sum of all data values divided by the total number of data values
Number of families in Neighborhood Q = 9
Mean family size of Neighborhood Q:
= (2 + 5 + 4 + 3 + 2 + 5 + 3 + 6 + 5) ÷ 9
= 35 ÷ 9
= 3.888888...
Number of families in Neighborhood S = 9
Mean family size of Neighborhood S:
= (2 + 3 + 2 + 3 + 7 + 2 + 3 + 3 + 2) ÷ 9
= 27 ÷ 9
= 3
The mean family size of Neighborhood Q is 3.88.. and the mean family size of Neighborhood S is 3. Therefore, Neighborhood Q appears to have a bigger family size as it's average family size is bigger than that of Neighborhood S.
Answer:
100 in^2
Step-by-step explanation:
Solve Area first:
5x8
= 40
Multiply by 2.5 (Enlarge)
40 x 2.5
= 100
Answer:
6.871 decimeters
Step-by-step explanation:
687.1 square centimeters = 6.871 square decimeters
1 dm² = 100 cm²
Answer: ( 77.27 , 83.93)
Therefore at 95% confidence/prediction interval is
= ( 77.27 , 83.93)
Step-by-step explanation:
Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.
The confidence interval of a statistical data can be written as.
x+/-zr/√n
Given that;
Mean x = 80.6 words per minute
Standard deviation r = 7.2
Number of samples n = 18
Confidence interval = 95%
z(at 95% confidence) = 1.96
Substituting the values we have;
80.6+/-1.96(7.2/√18)
80.6+/-1.96(1.697056274847)
80.6 +/- 3.33
= ( 77.27 , 83.93)
Therefore at 95% confidence/prediction interval is
= ( 77.27 , 83.93)
