Answer:
0.005 `; 0.00499 ;
No, because np < 10 ;
2000
Step-by-step explanation:
Given that:
Number of samples , n = 100
Proportion, p = x / n
p = 1 / 200
= 0.005
p = μ
Standard deviation of sample proportion :
σp = sqrt((p(1 - p)) / n)
σp = sqrt((0.005(1 - 0.005)) / 200)
σp = sqrt((0.005(0.995)) / 200)
σp = sqrt(0.004975 / 200)
σp = sqrt(0.000024875)
σp = 0.0049874
σp = 0.00499
np = 100 * 0.005 = 0.5
n(1 - p) = 100(1-0.05) = 95
Smallest value of n for which sampling distribution is approximately normal
np ≥ 10
0.005n ≥ 10
To obtain the smallest value of n,
0.005n = 10
n = 10 / 0.005
n = 2000
Hope this helps! I though writing everything down on paper would be more beneficial
Answer:
- 7 ; 7 pounds loss in weight.
Step-by-step explanation:
Given that :
Week ______ weight loss/ gain
1 ____________ - 2
2 ____________ - 6
3 ____________ + 7
4 ____________ - 6
Loss in weight is denoted by a negative sign;
The change in weight is the net sum of weight loss over the 4 weeks period
(-2) + (-6) + (7) + (-6)
-2 - 6 + 7 - 6 =-7
Hence, there is a 7 pounds loss in weight over the 4 weeks period.
Answer: the second option I think
Step-by-step explanation: