- El resultado de la conversión de 350 km/h a m/s es:

Suppose that at a state college, a random sample of 41 students is drawn, and each of the 41 students in the sample is asked to

ra1l [238]

Answer:

The confidence interval for 90% confidence would be narrower than the 95% confidence

Step-by-step explanation:

From the question we are told that

The sample size is n = 41

For a 95% confidence the level of significance is $\alpha = [100 - 95]\% = 0.05$ and

the critical value of $\frac{\alpha }{2}$ is $Z_{\frac{\alpha }{2} } =Z_{\frac{0.05 }{2} }= 1.96$

For a 90% confidence the level of significance is $\alpha = [100 - 90]\% = 0.10$ and

the critical value of $\frac{\alpha }{2}$ is $Z_{\frac{\alpha }{2} } =Z_{\frac{0.10 }{2} }= 1.645$

So we see with decreasing confidence level the critical value decrease

Now the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{n} }$

given that other values are constant and only $Z_{\frac{\alpha }{2} }$ is varying we have that

$E\ \ \alpha \ \ Z_{\frac{\alpha }{2} }$

Hence for reducing confidence level the margin of error will be reducing

The confidence interval is mathematically represented as

$\= x - E < \mu < \= x + E$

Now looking at the above formula and information that we have deduced so far we can infer that as the confidence level reduces , the critical value reduces, the margin of error reduces and the confidence interval becomes narrower

3 0

3 years ago

For a given IQ test, an individual is considered a genius if their score falls more than three standard deviations from the mean

ki77a [65]

Answer:

$P(Z>3) = 1-P(Z$

So then the probability that an individual present and IQ higher than 3 deviation from the mean is 0.00135

And if we find the number of individuals that can be considered as genius we got: 0.00135*1500=2.025

And we can say that the answer is a.2

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

2) Solution to the problem

Let X the random variable that represent the IQ scores of a population, and for this case we know the distribution for X is given by:

$X \sim N(\mu=?,\sigma=?)$