1. A GE lightbulb is suppose to last for 1,200 hours. in fact, light bulbs of this type last only 1,185 hours with a standard deviation of 70 hours. What is the probability that a sample of 100 lightbulbs will have an average life of at least 1,200 hours?
2. If a person rolls a sum of 8 when she rolls two dice, she wins $10. if the game is to be fair, how much (to the nearest penny) should the person pay to play the game?
1. The population standard deviation is given. For the sample standard deviation: s = σ ÷ √n = 70 ÷ √100 = 7. Let X be the number of hours a lightbulb lasts. Convert to z-scores:
z = (1200 – 1185) / 7 = 2.143
P(X ≥ 1200) = P(z ≥ 2.143) = normalcdf(2.143,99,0,1) = 0.0161
2. For the game to be fair, the dealer of the game must break even, that is, neither profit or lose any money. When two dice are rolled together, only 5 results will give a sum of 8. Therefore, the probability of a player winning is 5/36, or 0.13888.
Players receive $10 for a win; therefore, the break-even point (and thus, the amount each player should pay) is:
$10 × (5 ÷ 36) = $1.39