Rachel types 162 words in 3 minutes how many words did rachel type pure minute

In a random sample of students who took the SAT test, 427 had paid for coaching courses and the remaining 2733 had not. Calculat

Dvinal [7]

Answer:

The 95% confidence interval for the proportion of students who get coaching on the SAT is (0.1232, 0.147).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

427 had paid for coaching courses and the remaining 2733 had not.

This means that $n = 427 + 2733 = 3160, \pi = \frac{427}{3160} = 0.1351$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1351 - 1.96\sqrt{\frac{0.1351*0.8649}{3160}} = 0.1232$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1351 + 1.96\sqrt{\frac{0.1351*0.8649}{3160}} = 0.147$

The 95% confidence interval for the proportion of students who get coaching on the SAT is (0.1232, 0.147).

8 0

3 years ago

Let x denote the lifetime of a mcchine component with an exponential distribution. The mean time for the component failure is 25

aliina [53]

Answer:

0.1353 = 13.53% probability that the lifetime exceeds the mean time by more than 1 standard deviations

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The mean time for the component failure is 2500 hours.

This means that $m = \frac{2500}, \mu = \frac{1}{2500} = 0.0004$

What is the probability that the lifetime exceeds the mean time by more than 1 standard deviations?

The standard deviation of the exponential distribution is the same as the mean, so this is P(X > 5000).

$P(X > x) = e^{-0.0004*5000} = 0.1353$

0.1353 = 13.53% probability that the lifetime exceeds the mean time by more than 1 standard deviations

4 0

2 years ago

What is 66% of $1,678.89 plus 7.5% tax?

schepotkina [342]

50% of 1,678.89 = 839.445
10% = 167.889
5% = 83.9445
1% = 16.7889
.5% = 8.39445
Therefore 66% = 1108.0674
and 7.5% = 125.91675
So the two added together is: 1233.98415
To get the percentages simply do the following
50% = divide by two
10% = divide 50% by five OR divide the original number by 10
5% = divide 10% by two or divide 50% by 10
1% = divide 5% by 5 or divide original by 100
.5% = divide 1% by two

4 0

3 years ago

Answer this please. Thank you! Worth 10 points as well.

nlexa [21]

The probability that the next 4 winners will all be 7th grade students is 3/4

3 0

3 years ago

It was estimated that 650 people would attend bowling night, but 676 people actually attended.

Cerrena [4.2K]

Simply divide 650/676=.9615 or 96.15%. So there was 3.85% error

4 0

2 years ago