Question

A researches measure the lengths of movies and determines the mean length of a film is 115 minutes with a standard deviation of 11 minutes. What is the probability that a random chosen movie is greater than 120 minutes in length?

Answer 1

Solution: Given that mean = 115, sd = 11,

P(X > 120) = P((X-mean)/sd > (120-114)/11)
= P(Z > 0.5455)
= 1 - P(Z < 0.5455)
= 1 - 0.7088
= 0.2912

Z	0.00	0.01	0.02	0.03	0.04	0.05	0.06	0.07	0.08	0.09
0.0	0.5	0.504	0.508	0.512	0.516	0.5199	0.5239	0.5279	0.5319	0.5359
0.1	0.5398	0.5438	0.5478	0.5517	0.5557	0.5596	0.5636	0.5675	0.5714	0.5753
0.2	0.5793	0.5832	0.5871	0.591	0.5948	0.5987	0.6026	0.6064	0.6103	0.6141
0.3	0.6179	0.6217	0.6255	0.6293	0.6331	0.6368	0.6406	0.6443	0.648	0.6517
0.4	0.6554	0.6591	0.6628	0.6664	0.67	0.6736	0.6772	0.6808	0.6844	0.6879
0.5	0.6915	0.695	0.6985	0.7019	0.7054	0.7088	0.7123	0.7157	0.719	0.7224