First write the equation
3/n = 5
Multiply both sides by n
3/n * n = 5 * n
Simplify
3 = 5 * n
Divide both sides by 5
3/5 = 5 * n/5
Simplify
3/5 = n
Divide
0.6 = n
So 3/0.6 = 5
N=25000, p=0.001, np=25000*0.001=25 >10, so the normal approximation should apply.
fewer than 20 => 0,1,2,3...19
Applying the continuity correction, we have X<19.5.
μ=np=25
σ²=npq=25000*.001*.999=24.975 =>
σ=√(24.975)=4.9975
Z(X<19.5)=(19.5-25)/4.9975=-1.100550
P(X<19.5)=P(Z<-1.10055)=0.1355463 [from normal probability table, left tail]
Hence normal approximation to probability that fewer than 20 cars will be stolen is 0.1355.
[actual value by binomial distribution is 0.133450. approximation is quite close]
15-6.9=8.1
8.1 x 3.9= 31.59
$31.59 is your answer
hope this helps :)
Distance = rate * time (I will abbreviate as d = rt)
d = rt, so r = d/t
r = 20mi/24min
= 0.833333 mi/min.
d = rt again for the new time
d = 0.8333333 mi/min * 6 min = 5 miles
