Answer:
a.![\mu=15](https://tex.z-dn.net/?f=%5Cmu%3D15)
b.![\mu=7.8586 \ and \ \mu=22.1414](https://tex.z-dn.net/?f=%5Cmu%3D7.8586%20%5C%20and%20%20%5C%20%5Cmu%3D22.1414)
c. Choice A- Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.
Step-by-step explanation:
a.Binomial distribution is defined by the expression
![P(X=k)=C_k^n.p^k.(1-p)^{n-k}](https://tex.z-dn.net/?f=P%28X%3Dk%29%3DC_k%5En.p%5Ek.%281-p%29%5E%7Bn-k%7D)
Let n be the number of trials,![n=100](https://tex.z-dn.net/?f=n%3D100)
and p be the probability of success,![p=15\%](https://tex.z-dn.net/?f=p%3D15%5C%25)
The mean of a binomial distribution is the probability x sample size.
![\mu=np=100\times0.15=15](https://tex.z-dn.net/?f=%5Cmu%3Dnp%3D100%5Ctimes0.15%3D15)
b.Limits within which p is approximately 95%
sd of a binomial distribution is given as:![\sigma=\sqrt npq\\q=1-p](https://tex.z-dn.net/?f=%5Csigma%3D%5Csqrt%20npq%5C%5Cq%3D1-p)
Therefore, ![\sigma=\sqrt(100\times0.015\times0.85)=3.5707](https://tex.z-dn.net/?f=%5Csigma%3D%5Csqrt%28100%5Ctimes0.015%5Ctimes0.85%29%3D3.5707)
Use the empirical rule to find the limits. From the rule, approximately 95% of the observations are within to standard deviations from mean.
![sd_1=>\mu-2\sigma=15-2\times3.3507=7.8586\\sd_2=>\mu-2\sigma=15+2\times3.3507=22.1414](https://tex.z-dn.net/?f=sd_1%3D%3E%5Cmu-2%5Csigma%3D15-2%5Ctimes3.3507%3D7.8586%5C%5Csd_2%3D%3E%5Cmu-2%5Csigma%3D15%2B2%5Ctimes3.3507%3D22.1414)
Hence, approximately 95% of the observations are within 7.8586 and 22.1414 (areas of infestation).
c.
is not within the limits in b above (7.8586,22.1414). X=45 appears to be a large area of infestation. A.Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.