Percent of red lights last between 2.5 and 3.5 minutes is 95.44% .
<u>Step-by-step explanation:</u>
Step 1: Sketch the curve.
The probability that 2.5<X<3.5 is equal to the blue area under the curve.
Step 2:
Since μ=3 and σ=0.25 we have:
P ( 2.5 < X < 3.5 ) =P ( 2.5−3 < X−μ < 3.5−3 )
⇒ P ( (2.5−3)/0.25 < (X−μ)/σ < (3.5−3)/0.25)
Since, Z = (x−μ)/σ , (2.5−3)/0.25 = −2 and (3.5−3)/0.25 = 2 we have:
P ( 2.5<X<3.5 )=P ( −2<Z<2 )
Step 3: Use the standard normal table to conclude that:
P ( −2<Z<2 )=0.9544
Percent of red lights last between 2.5 and 3.5 minutes is 
% .
Answer:
Sammy charges less by $3.75
Step-by-step explanation:
givens
-Sammy: 21.75 for 3 hrs
-y=8x for Claire
-5 hours, and who charges less?
21.75/3
$7.25/hr
y=8(5)
y=$40
y=7.25(5)
y=$36.25
40-36.25
$3.75
The correct option is B.
Pan balance is one of the instruments that are used to measure weights of objects. The instrument is designed in such a way that, it has two pans and a standard weight can be added to one of the pan to determine the weight of the object in the other pan. In the question given above, the pan balance is to be used to measure the weight of a large rock. Salt is the right object to be used on the second pan because it can be packaged in standard weights and it is very light in weight. So, it will be very sensitive to the weight of the rock that is been added on the other pan.
Situation satisfies the criteria for the use of hypergeometric distribution. Since no replacement is made, binomial distribution is not applicable (probability does not remain constant).
A=number of target wattage bulbs
B=number of non-targeted wattage bulbs
a=number of target wattage bulbs selected
b=number of non-targeted wattage bulbs selected
P(a,b)=C(A,a)*C(B,b)/C(A+B,a+b)
where C(n,x)=combination of x items chosen from n=n!/(x!(n-x)!)
For all following problems,
A+B=4+5+6=15
a+b=3 (selected)
(a) Target wattage = 75W
A=6, B=9, a=2, b=1
P(a,b,A,B)
=P(2,1,6,9)
=C(6,2)*C(9,1)/C(15,3)
=15*9/455
=27/91
(b) target wattage = each of the three
Probability = sum of probabilities of choosing 3 40,60,75-watt bulbs
P(3x40W)+P(3x60W)+P(3x75W)
Case (A,B,a,b)
3x40W (4,11,3,0)
3x60W (5,10,3,0)
3x75W(6,9,3,0)
P(3x40W)+P(3x60W)+P(3x75W)
=C(4,3)*C(11,0)/C(15,3)+C(5,3)*C(10,0)/C(15,3)+C(6,3)*C(9,0)/C(15,3)
=4*1/455+10*1/455+20*1/455
=34/455
Can also be solved by elementary counting, for example, for (a),
P(2x75W)
=C(3,2)*6/15*5/14*9/13
=(3)*6/15*5/14*9/13
=27/91 as before
