Answer: a. 0.93449
b. 0.99506
Step-by-step explanation:
Given : The proportion of all drivers in a certain state regularly wear a seat belt : p=0.75
Sample size : n=500
Let x be the random variable that represents the number of drivers in a certain state regularly wear a seat belt.
Then , [tex]\mu=np=(500)(0.75)=375[/tex]
[tex]\sigma=\sqrt{np(1-p)}=\sqrt{500)(0.75)(0.25)}=9.68[/tex]
a. The probability that between 360 and 400 (inclusive) of the drivers in the sample regularly wear a seat belt :
[tex]P(360<x<400)=P(\dfrac{360-375}{9.68}<z<\dfrac{400-375}{9.68})\ \ [\because z=\dfrac{x-\mu}{\sigma}]\\\\=P(-1.55<z<2.58)\\\\=P(z<2.58)-P(z<-1.55)\\\\=P(z<2.58)-(1-P(z<1.55))\ \ [\because P(Z<-z)=1-P(Z<z)][/tex]
[tex]=0.99506-(1- 0.9394292)\\\\=0.9344892\approx0.93449[/tex]
∴ The probability that between 360 and 400 (inclusive) of the drivers in the sample regularly wear a seat belt= 0.93449
b. The probability that fewer than 400 of those in the sample regularly wear a seat belt :-
[tex]P(x<400)=P(\dfrac{x-\mu}{\sigma}<\dfrac{400-375}{9.68})\\\\=P(z<2.58)=0.99506[/tex]
∴ The probability that fewer than 400 of those in the sample regularly wear a seat belt = 0.99506
