Scores on a university exam are Normally distributed with a mean of 78 and a standard deviation of 8. The professor teaching the class declares that a score of 62 or higher is required for a grade of at least a D. Using the 68-95-99.7 rule, what percentage of students score below 62

Porcentage of students score below 62 is close to 0,08%

Step-by-step explanation:

The rule

68-95-99.7

establishes:

The intervals:

[ μ₀ - 0,5σ ,  μ₀ + 0,5σ] contains 68.3 % of all the values of the population

[ μ₀ - σ ,  μ₀ + σ]   contains 95.4 % of all the values of the population

[ μ₀ - 1,5σ ,  μ₀ + 1,5σ] contains 99.7 % of all the values of the population

In our case such intervals become

[ μ₀ - 0,5σ ,  μ₀ + 0,5σ]   ⇒  [ 78 - (0,5)*8 , 78 + (0,5)*8 ]  ⇒[ 74 , 82]

[ μ₀ - σ ,  μ₀ + σ]  ⇒ [ 78 - 8 , 78 +8 ]   ⇒  [ 70 , 86 ]

[ μ₀ - 1,5σ ,  μ₀ + 1,5 σ]  ⇒ [ 78 - 12 , 78 + 12 ]  ⇒ [ 66 , 90 ]

Therefore the last interval

[ μ₀ - 1,5σ ,  μ₀ + 1,5 σ]    ⇒  [ 66 , 90 ]

has as lower limit 66 and contains 99.7 % of population, according to that the porcentage of students score below 62 is very small, minor than 0,15 %

100 - 99,7  = 0,3 %

Only 0,3 % of population is out of   μ₀ ± 1,5 σ, and by symmetry 0,3 /2 = 0,15 % is below the lower limit, 62 is even far from 66 so we can estimate, that the porcentage of students score below 62 is under 0,08 %

a) $ 9.50

b) 17

c) y-intercept (0, 12). this is the initial amount of money that joe had prior to playing any game.

d) x-intercept ( 24, 0). this is the number of games that joe will play before he depletes the initial amount of money that he had.

e) slope is -0.5. joe's balance will reduce by 0.5 for every game that he plays.

f)

domain; [0,24]

range; [0,12]

g) x + 2y = 24

step-by-step explanation:

a)

if joe plays 5 games, this will imply that x = 5. substitute x = 5 in the equation and solve for y;

y = -.50x + 12

y = -0.50(5) + 12

y = 9.50

b)

if joe leaves the arcade with $3.50, this will mean that y = 3.50. plug in y = 3.50 in the given equation and solve for x;

y = -.50x + 12

3.50 = -0.50x + 12

0.50x = 12 - 3.50

0.5x = 8.5

x = 17

c)

the function given is   y = -.50x + 12.

the y-intercept is the point where the graph of this function will intersect the y-axis. at this point, the value of x is usually 0. plug in x = 0 in the given equation and determine y;

y = -0.50(0) + 12

y = 12

the y-intercept is thus (0, 12). this is the initial amount of money that joe had prior to playing any game.

d)

the x-intercept refers to the point where the graph of a function intersects with the x-axis. at this point, the value of y is 0. substitute y = 0 in the given equation and solve for x;

0 =   -.50x + 12

0.5x = 12

x = 24

x-intercept ( 24, 0). this is the number of games that joe will play before he depletes the initial amount of money that he had.

e)

the slope of the given equation is -0.5. the equation has been given in slope-intercept form. thus the slope is simply the coefficient of x.

joe's balance will reduce by 0.5 for every game that he plays. this is the interpretation of the slope.

f)

domain is the set of x values for which the function is real and defined. the domain will thus be; [0,24]. this will restrict the balance to only assume positive values.

on the other hand, the range is the set of y values for which a function is real and defined. the range in our case will be [0,12]. joe can have between $0 and $12 balance.

g)

the standard form of the equation of a straight line is;

ax + by = c

where a, b, and c are integers.

our equation is given as;

y = -.50x + 12

re-writing the equation we have;

0.5x + y = 12

divide all through by 0.5;

x + 2y = 24