Given that mean=3750 hours and standard deviation is 300:
Then:
<span>a. The probability that a lamp will last for more than 4,000 hours?
P(x>4000)=1-P(x<4000)
but
P(x<4000)=P(z<Z)
where:
z=(x-</span>μ)/σ
z=(4000-3750)/300
z=0.833333
thus
P(x<4000)=P(z<0.8333)=0.7967
thus
P(x>4000)=1-0.7967=0.2033
<span>b.What is the probability that a lamp will last less than 3,000 hours?
P(x<3000)=P(z<Z)
Z=(3000-3750)/300
z=-2.5
thus
P(x<3000)=P(z<-2.5)=0.0062
c. </span><span>.What lifetime should the manufacturer advertise for these lamps in order that only 4% of the lamps will burn out before the advertised lifetime?
the life time will be found as follows:
let the value be x
the value of z corresponding to 0.04 is z=-2.65
thus
using the formula for z-score:
-2.65=(x-3750)/300
solving for x we get:
-750=x-3750
x=-750+3750
x=3000</span>
Answer:
start with root and minus with 2 and 5 and answer 16b
Answer:
25.133 units
Step-by-step explanation:
Since the density ρ = r, our mass is
m = ∫∫∫r³sinθdΦdrdθ. We integrate from θ = 0 to π (since it is a hemisphere), Φ = 0 to 2π and r = 0 to 2 and the maximum values of r = 2 in those directions. So
m =∫∫[∫r³sinθdΦ]drdθ
m = ∫[∫2πr³sinθdθ]dr ∫dФ = 2π
m = ∫2πr³∫sinθdθ]dr
m = 2π∫r³dr ∫sinθdθ = 1
m = 2π × 4 ∫r³dr = 4
m = 8π units
m = 25.133 units
