The mean and standard deviation of a normal distribution are 150 and 12 respectively. What percentage of values lie between:
a) 138 and 162
b) 126 and 174
c) 126 and 162
d) 162 and 174

a) (150 - 138) / 12 = 1 deviation or 34%

(162 - 150) / 12 = 1 deviation or 34%

34% + 34% = 68%

b) (150 - 126) / 12 = 2 deviations or 47.5%

(174 - 150) / 12 = 2 deviations or 47.5%

47.5% + 47.5% =95%

c) (150 - 126) / 12 = 2 deviations or 47.5%

(162 - 150) / 12 = 1 deviation or 34%

47.5% + 34% = 81.5%

d) (162 - 150) / 12 = 1 deviation or 34%

(174 - 150) / 12 = 2 deviations or 47.5%

This problem is a little different. We have to subtract 34% from 47.5%

47.5% - 34% = 13.5%

1. sometimes

2.   never

3   .08c = 7.50+.05c   250 pages

step-by-step explanation:

1.   sometimes

the theoretical probability of rolling a 1 is 1/6.   perhaps you got really lucky and never rolled a 1   in those 60 times.   your experimental probability is 0.   or perhaps you rolled a 1   20 times.   your experimental probability is 20/60 is 1/3.     it really just depends on how you rolled in that experiment.

2.   never

the probability of the event occurring is 1.   that is being 100 % sure it will happen.   that is the maximum amount of confidence we have.

3.   no customer card cost:   .08 times number of pages

customer card cost:   7.50 + .05 times number of pages

we want to find when it costs the same amount, so we will set these equal

let c = number of pages

.08c = 7.50+.05c

subtract .05c from each side

.08c-.05c = 7.5+.05c-.05c

.03c=7.5

divide each side by .03

.03c/.03 = 7.5/03

c =250  

the cost is the same at 250 pages

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and  

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

:

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

:

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)