A swimming pool has an average depth of 1.6 m. It is 50 m long and 30 m wide. What is the volume?

A high school principal wishes to estimate how well his students are doing in math. Using 40 randomly chosen tests, he finds tha

ollegr [7]

Answer:

99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].

Step-by-step explanation:

We are given that a high school principal wishes to estimate how well his students are doing in math.

Using 40 randomly chosen tests, he finds that 77% of them received a passing grade.

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of students received a passing grade = 77%

n = sample of tests = 40

p = population proportion

Here for constructing 99% confidence interval we have used One-sample z proportion test statistics.

So, 99% confidence interval for the population proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5%

level of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [ $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.77-2.5758 \times {\sqrt{\frac{0.77(1-0.77)}{40} } }$ , $0.77+2.5758 \times {\sqrt{\frac{0.77(1-0.77)}{40} } }$ ]

= [0.5986 , 0.9414]

Therefore, 99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].

Lower bound of interval = 0.5986

Upper bound of interval = 0.9414

6 0

3 years ago

Solve the Equation 2x-3=14 -x+3y=-6

andriy [413]

Step-by-step explanation:

If you lost "y" in the first equation.

$\underline{+\left\{\begin{array}{ccc}2x-3y=14\\-x+3y=-6\end{array}\right}\qquad\text{add both sides of the equations}\\\\.\qquad x=8\\\\\text{Put the value of x to the first equation:}\\\\2(8)-3y=14\\16-3y=14\qquad\text{subtract 16 from both sides}\\-3y=-2\qquad\text{divide both sides by (-3)}\\y=\dfrac{2}{3}\\\boxed{x=8,\ y=\dfrac{2}{3}}$

If first equation is correct.

$2x-3=14\qquad\text{add 3 to both sides}\\2x=17\qquad\text{divide both sides by 2}\\x=8.5\\\\\text{Put the value of x to the second equation:}\\-8.5+3y=-6\qquad\text{add 8.5 to both sides}\\3y=2.5\qquad\text{divide both sides by 3}\\y=\dfrac{2.5}{3}\\y=\dfrac{25}{30}\\y=\dfrac{5}{6}\\\\\boxed{x=8.5,\ y=\dfrac{5}{6}}$