Information on a packet of seeds claims that the germination rate is 92%. The packet contains 160 seeds. Let p represent the proportion of seeds in the packet that will germinate.The sampling distribution model for p is:a. N(0.92, 0.0215)b. N(0.08, 0.0215)c. N(147.2, 12.8)d. N(0.0215, 0.92)

Answer: a. N(0.92, 0.0215)Step-by-step explanation:For population proportion (p) and sample size n, the mean and standard deviation is given by :-The sampling distribution model for p is given by :_[tex]N(\mu_p,\ \sigma_p)[/tex], where [tex]\mu_p=p \\ \sigma_p=\sqrt{\dfrac{p(1-p)}{n}[/tex]We assume that the seeds are randomly selected.Given : Information on a packet of seeds claims that the germination rate is 92%.i.e. p= 0.92The packet contains 160 seeds.i.e. n= 160Then ,  [tex]\mu_p=0.92\\ \sigma_p=\sqrt{\dfrac{0.92(1-0.92)}{160}}\approx0.0215[/tex]Hence, the sampling distribution model for p is: [tex] N(\mu_p,\sigma_p) = N(0.92, 0.0215) [/tex]