To evaluate the probability that a randomly selected day will be between 28 and 34 minutes we proceed as follows:
P(28<x<34)
First we evaluate the z-score for the above values:
z=(x-σ)/μ
μ=26.7
σ=5.1
when:
x=28
z=(28-26.7)/5.1
z=0.26
P(z<0.26)=0.6026
when x=34
z=(34-26.7)/5.1
z=1.43
P(z<1.43)=0.9236
hence:
P(28<x<34)=0.9236-0.6026=0.321~32.1%
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:
A. 102
Step-by-step explanation:
102/6 = 17
611/6 = 101.83333....
613/6 = 102.1666...
614/6 = 103.3333......
Answer:
m = 4/3
Step-by-step explanation:
