A certain brand of flood lamps has a lifetime that is normally distributed with a mean of 3,750 hours and a standard deviation of 300 hours.

(a) What proportion of these lamps will last for at least 4,000 hours?

(b) What proportion of these lamps will last between 3,600 and 4,000 hours?

(c) What proportion of these lamps will last less than 3,800?

0.83

Step-by-step explanation:

(a) z = (x - mean) / sd = (4000 - 3750) / 300 = 0.83

This corresponds to 0.7967 from the z-table, which represents how many last less than 4000 hours, so the answer is 1 - 0.7967 = 0.2033

(b) It will be P(x<4000) - P(x>3600) = 0.7967 - whatever the corresponding value is.

(c) same procedure as (b).

(d) 2% above corresponds to 0.02 which is a z-score of -2.05

-2.05 = (x - 3750) / 300

x = 3135

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18

step-by-step explanation: