Answer:
0.1052
Step-by-step explanation:
Given that proportion of germination in the population is 94% =0.94
p = 0.96
Sample size = 220
Std dev of p = [tex]\sqrt{\frac{pq}{n} } \\=0.016[/tex]
The probability that more than 96% of the 220 seeds in the packet will germinate
= [tex]P(p\geq 0.96)\\=P(Z\geq \frac{0.96-0.94}{0.016})\\=P(Z\geq 1.25)\\==1-0.8948\\=0.1052[/tex]
Assumptions are np and nq >5 and also sample size >220 hence normal
