A weather forecaster predicts that there is a 45% probability of snow today. What is the probability that it will not snow?

A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 46.046.0 and

Gala2k [10]

Answer:

The probability of selecting a class that runs between 50.2550.25 and 51.2551.25 minutes is 0.10

Step-by-step explanation:

The Uniform Distribution, also known as Rectangular Distribution, is a type of Continuous Probability Distribution. It has a continuous random variable restricted to a finite interval and its probability function has a constant density during this interval.

The formula of probability if given by:

f(x)=

$\left \{ {{\frac{1}{b-a}; \ a \leq x \leq b } \atop {0}; \ x \ otherwise } \right.$

In this exercise a= 46.0 and b= 56.0

The probability of selecting a class that runs between 50.2550.25 and 51.2551.25 minutes is:

$\int\limits^{51.25}_{50.25} {\frac{1}{56-46} } \, dx = \int\limits^{51.25}_{50.25} {\frac{1}{10} } \, dx = \frac{1}{10} \times (51.25 - 50.25)=\frac{1}{10}=0.1$

7 0

3 years ago

5. Find the squares of the following numbers by using the formula of (a + b)' or (a-b]². c) 98

Dmitrij [34]

98^2 = (100-2)^2

Using (a-b)^2, we have

= (100)^2-2(100)(2)+(2)^2

= 10000-400+4

= 10004-400

= 9604

The answer is 9604.

Please mark me Brainliest

5 0

3 years ago

The life of a manufacturer's compact fluorescent light bulbs is normal, with mean 12,000 hours and standard deviation 2,000 hour

vitfil [10]

Z value is a numerical measurement that describe a value relationship to the mean of a group of values. The standard deviations is 1.25 above the mean is 14,500 hours.

<h3>Given information-</h3>

The mean for the bulb is 12,000 hours.

The standard deviation for the bulb is 2000 hours.

Sample value is 14500.

To find out the how many standard deviation is 14500 mean away from the mean the z value of the mean should be calculated.

<h3>Z value</h3>

Z value is a numerical measurement that describe a value relationship to the mean of a group of values. Z value is the ratio of the difference of the sample value $x$ and mean $\mu$ to the standard deviation. Thus the z value for the given mean $\sigma$ is,

$z=\dfrac{x-\mu}{\sigma}$

$z=\dfrac{14500-1200}{2000}$

$z=1.25$

Thus the standard deviations is 1.25 above the mean is 14,500 hours.

Learn more about the z value here;

brainly.com/question/62233

4 0

2 years ago

help............................................................................................................................

dlinn [17]

Answer:6/25

Step-by-step explanation:

6 0

3 years ago

Suppose that a local TV station conducts a survey of a random sample of 120 registered voters in order to predict the winner of

forsale [732]

Answer:

a) The 99% CI for the true proportion of voters who prefer the Republican candidate is (0.3658, 0.6001). This means that we are 99% sure that the true population proportion of all voters who prefer the Republican candidate is (0.3658, 0.6001).

b) The upper bound of the confidence interval is above 0.5 = 50%, which meas that the candidate can be confidence of victory.

Step-by-step explanation:

Question a:

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 $1 - \frac{\alpha}{2}$ .

Sample of 120 registered voters in order to predict the winner of a local election. The Democrat candidate was favored by 62 of the respondents.

So 120 - 62 = 58 favored the Republican candidate, so:

$n = 120, \pi = \frac{58}{120} = 0.4833$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a p-value of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4833 - 2.575\sqrt{\frac{0.4833*0.5167}{120}} = 0.3658$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4833 + 2.575\sqrt{\frac{0.4833*0.5167}{120}} = 0.6001$

The 99% CI for the true proportion of voters who prefer the Republican candidate is (0.3658, 0.6001). This means that we are 99% sure that the true population proportion of all voters who prefer the Republican candidate is (0.3658, 0.6001).

b. If a candidate needs a simple majority of the votes to win the election, can the Republican candidate be confident of victory? Justify your response with an appropriate statistical argument.

The upper bound of the confidence interval is above 0.5 = 50%, which meas that the candidate can be confidence of victory.

8 0

3 years ago