Respuesta :
Answer:
Probability that 24 or more of those drivers were involved in accidents is 0.15625.
Step-by-step explanation:
We are given that an insurance agent research suggested that first year drivers had roughly an 13% chance of being involved in an automobile accident while driving.
The insurance agent provided insurance to 152 first year drivers last year.
Let [tex]\hat p[/tex] = sample proportion of drivers who were involved in accidents
The z score probability distribution for sample proportion is given by;
Z = [tex]\frac{\hat p-p}{\sqrt\frac{\hat p(1-\hat p)}{n} {} }[/tex] ~ N(0,1)
where, p = population proportion of first year drivers involved in an automobile accident while driving = 13%
n = sample of first year drivers = 152
Now, probability that 24 or more of those drivers were involved in accidents is given by = P([tex]\hat p[/tex] [tex]\geq[/tex] [tex]\frac{24}{152}[/tex])
P([tex]\hat p[/tex] [tex]\geq[/tex] 0.16) = P( [tex]\frac{\hat p-p}{\sqrt\frac{\hat p(1-\hat p)}{n} {} }[/tex] [tex]\geq[/tex] [tex]\frac{0.16-0.13}{\sqrt\frac{0.16(1-0.16)}{152} {} }[/tex] ) = P(Z [tex]\geq[/tex] 1.01) = 1 - P(Z < 1.01)
= 1 - 0.84375 = 0.15625
The above probability is calculated by looking at the value of x = 1.01 in the z table which has an area of 0.84375.
Hence, the required probability is 0.15625.
