The most reasonable answer is b
The z-score for 5 minutes is found from:
![z=\frac{X-\mu}{\sigma} ](https://tex.z-dn.net/?f=z%3D%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%0A)
![z=\frac{5-4}{0.6}=1.667](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B5-4%7D%7B0.6%7D%3D1.667)
Reference to a normal distribution table shows that the cumulative probability for a z-score of 1.667 is around 0.9522.
Therefore the probability that a conversation lasts for longer than 5 minutes is:
1 - 0.9522 = 0.048 to 3 d.p.
Answer:
b) 65.08
ii) 0.75
Step-by-step explanation:
x(Kg): 62 63 64 65 66 67 68
Frequency(f): 1 5 11 15 10 6 2
xf: 62 315 704 975 660 402 136
b) mean(X) = ∑xf/N = 3254/50 = 65.08
c) x - X : -3.08 -2.08 -1.08 -0.08 0.92 1.92 2.92
(x - X)²: 9.49 4.33 1.17 0.0064 0.85 3.69 8.53
ii) Standard deviation(S.D.): √(∑(x - X)²/N = √(28.0664)/50 = √0.5613
∴ S.D.= 0.75
Answer:
1 ad 12
Step-by-step explanation:
Ok so t=1.95p for each cost of pound