Answer:
The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447
Explanation:
Probability of wet weather = 0.15
Probability of not being a wet weather = 1-0.15
We are supposed to find probability that it will take a week for it three wet weather on 3 separate days
Total number of days in a week = 7
We will use binomial over here
n = 7
p =probability of failure = 0.15
q = probability of success=1-0.15
r=3
Formula :![P(r=3)=^nC_r p^r q ^{n-r}](https://tex.z-dn.net/?f=P%28r%3D3%29%3D%5EnC_r%20p%5Er%20q%20%5E%7Bn-r%7D)
![P(r=3)=^{7}C_{3} (0.15)^3 (1-0.15)^{7-3}\\P(r=3)=\frac{7!}{3!(7-3)!} (0.15)^3 (1-0.15)^{7-3}\\P(r=3)=0.06166](https://tex.z-dn.net/?f=P%28r%3D3%29%3D%5E%7B7%7DC_%7B3%7D%20%280.15%29%5E3%20%281-0.15%29%5E%7B7-3%7D%5C%5CP%28r%3D3%29%3D%5Cfrac%7B7%21%7D%7B3%21%287-3%29%21%7D%20%280.15%29%5E3%20%281-0.15%29%5E%7B7-3%7D%5C%5CP%28r%3D3%29%3D0.06166)
Standard deviation =![\sqrt{n \times p \times q}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%20%5Ctimes%20p%20%5Ctimes%20q%7D)
Standard deviation =![\sqrt{7 \times 0.15 \times (1-0.15)}](https://tex.z-dn.net/?f=%5Csqrt%7B7%20%5Ctimes%200.15%20%5Ctimes%20%281-0.15%29%7D)
Standard deviation =0.9447
Hence The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447
are you serious I see this only I see this so many days after
:P
Answer:
0.35
Explanation:
( 0.45*1 + 0.25*1 ) / 2 = 0.35
Answer:
yes
Explanation:
due to the cooling effect