Answer: 0.0594
Step-by-step explanation:
Given : The population proportion of adults questions reported that their health was excellent : p = 0.42
Among the 14 adults randomly selected from this area, only 3 reported that their health was excellent.
i.e. Sample size : n= 14
[tex]\mu=np = 14\times0.42=5.88[/tex]
[tex]s=\sqrt{np(1-p)}=\sqrt{14(0.42)(1-0.42)}\approx1.85[/tex]
z-score : [tex]z=\dfrac{{x}-\mu}{\sigma}[/tex]
Put x= 3
[tex]z=\dfrac{3-5.88}{1.85}\approx-1.56[/tex]
Then, the probability that only 3 or fewer report having excellent health:-
[tex]P(z\leq-1.56)=0.0593799\approx0.0594[/tex]
Hence, the required probability : 0.0594
