Question

The answers are in red. Please explain all, esp part d!

The average number of miles driven on a full tank of gas in a certain model car before its low-fuel light comes on is 331. Assume this mileage follows the normal distribution with a standard deviation of 34 miles. Complete parts a through d below. a. What is the probability that, before the low-fuel light comes on, the car will travel less than 349 miles on the next tank of gas? 0.7017 (Round to four decimal places as needed.) b. What is the probability that, before the low-fuel light comes on the car will travel more than 273 miles on the next tank of gas? 0.9560 (Round to four decimal places as needed.) c. What is the probability that, before the low-fuel light comes on, the car will travel between 271 and 291 miles on the next tank of gas? 0.0809 (Round to four decimal places as needed.) d. What is the probability that, before the low-fuel light comes on the car will travel exactly 303 miles on the next tank of gas? 0.0000 (Round to four decimal places as needed.)

Answer 1

a) Let X be the number of miles driven by the car before the low fuel light turns on

$X\sim N(331,34^2)$

then

$z= \frac{X-331 }{34}\sim N(0,1)$

$P(X<349)= P(z<\frac{349-331}{34})$

= P( z < 0.53)

= 0.7019 (from z table)

(b)

$P(X>273)= P(z>\frac{273-331}{34})$

= P( z > -1.71)

= 0.9564 (from z table)

(c)

$P(271<X<291)= P(\frac{271-331}{34}<z<\frac{291-331}{34})$

= P( -1.76 < z < -1.18 )

= P( -1.76 < z < 0) - P( -1.18 < z < 0)

= 0.4608 - 0.3810 (from z table)

= 0.0798

(d) Using continuity correction

P( X=303) = P( 302.5<X < 303.5)

= P( -0.84 < z < -0.81)

= 0.2995 - 0.2910   (from z table)

= 0.0085

X-N(331, 34)

X - 331 E N (0,1)

349-331 P(X <349) = P(Z <= 34

273 - 331 P(X > 273) = P(Z >= 34

271 - 331 P(271 < X < 291) = P(- 291 - 331 34 34