Question

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 1

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 :_

$N(\mu_p,\ \sigma_p)$

, where

$\mu_p=p \\ \sigma_p=\sqrt{\dfrac{p(1-p)}{n}$

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= 0.92

The packet contains 160 seeds.

i.e. n= 160

Then ,  $\mu_p=0.92\\ \sigma_p=\sqrt{\dfrac{0.92(1-0.92)}{160}}\approx0.0215$

Hence, the sampling distribution model for p is:

$N(\mu_p,\sigma_p) = N(0.92, 0.0215)$