231 is 75% of what number?

Scores on a university exam are Normally distributed with a mean of 78 and a standard deviation of 8. The professor teaching the

mafiozo [28]

Answer:

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 %

6 0

3 years ago

write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the zeros 2, -3, a

AysviL [449]

Answer:

Do you have some examples?

5 0

3 years ago

What is the solution to the equation One-fourth x minus one-eighth = StartFraction 7 Over 8 EndFraction + one-half x?

snow_tiger [21]

Answer:

The solution to the equation is

x = -4

Step-by-step explanation:

We want to find the solution to the equation

(1/4)x - 1/8 = 7/8 + (1/2)x

First, add -(1/2)x + 1/8 to both sides of the equation.

(1/4)x - 1/8 - (1/2)x + 1/8 = 7/8 + (1/2)x - (1/2)x + 1/8

[1/4 - 1/2]x = 7/8 + 1/8

x(1 - 2)/4 = (7 + 1)/8

-(1/4)x = 8/8

-(1/4)x = 1

Multiply both sides by -4

x = -4

3 0

4 years ago

Read 2 more answers

A university wants to compare out-of-state applicants' mean SAT math scores (?1) to in-state applicants' mean SAT math scores (?

nordsb [41]

Answer:

d. Yes, because the confidence interval does not contain zero.

Step-by-step explanation:

We are given that the university looks at 35 in-state applicants and 35 out-of-state applicants. The mean SAT math score for in-state applicants was 540, with a standard deviation of 20.

The mean SAT math score for out-of-state applicants was 555, with a standard deviation of 25.

Firstly, the Pivotal quantity for 95% confidence interval for the difference between the population means is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean SAT math score for in-state applicants = 540

$\bar X_2$ = sample mean SAT math score for out-of-state applicants = 555

$s_1$ = sample standard deviation for in-state applicants = 20

$s_2$ = sample standard deviation for out-of-state applicants = 25

$n_1$ = sample of in-state applicants = 35

$n_2$ = sample of out-of-state applicants = 35

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(35-1)\times 20^{2} +(35-1)\times 25^{2} }{35+35-2} }$ = 22.64

<em>Here for constructing 95% confidence interval we have used Two-sample t test statistics.</em>

So, 95% confidence interval for the difference between population means ( $\mu_1-\mu_2$ ) is ;

P(-1.997 < $t_6_8$ < 1.997) = 0.95 {As the critical value of t at 68 degree

of freedom are -1.997 & 1.997 with P = 2.5%}

P(-1.997 < $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 1.997) = 0.95

P( $-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ${(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}$ < $1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.95

P( $(\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.95

<u>95% confidence interval for</u> ( $\mu_1-\mu_2$ ) =

[ $(\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ]

=[ $(540-555)-1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } }$ , $(540-555)+1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } }$ ]

= [-25.81 , -4.19]

Therefore, 95% confidence interval for the difference between population means SAT math score for in-state and out-of-state applicants is [-25.81 , -4.19].

This means that the mean SAT math scores for in-state students and out-of-state students differ because the confidence interval does not contain zero.

So, option d is correct as Yes, because the confidence interval does not contain zero.

6 0

3 years ago

What is the area of the shape

Verizon [17]

The area of the shape is 180 square feet if the length of the diagonals are 20 and 18 ft respectively, option (a) is correct.

<h3>What is quadrilateral?</h3>

It is defined as the four-sided polygon in geometry having four edges and four corners. Two pairs of congruent sides, and it has one pair of opposite congruent angles.

We know the area of the kite = pq/2

Where p and q are the length of the diagonals

p = 10+10 = 20 ft

q = 7+11 = 18 ft

Area = (20×18)/2

Area = 180 square feet

Thus, the area of the shape is 180 square feet if the length of the diagonals are 20 and 18 ft respectively option (a) is correct.

Learn more about the quadrilateral here:

brainly.com/question/6321910

#SPJ1

3 0

2 years ago