When divide a fraction less than one by a whole number greater than one, how does the quotient compare to the dividend .

Suppose that the walking step lengths of adult males are normally distributed with a mean of 2.5 feet and a standard deviation o

Hitman42 [59]

Answer:

0.0668 = 6.68% probability that an individual man’s step length is less than 1.9 feet.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with a mean of 2.5 feet and a standard deviation of 0.4 feet.

This means that $\mu = 2.5, \sigma = 0.4$

Find the probability that an individual man’s step length is less than 1.9 feet.

This is the p-value of Z when X = 1.9. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1.9 - 2.5}{0.4}$

$Z = -1.5$

$Z = -1.5$ has a p-value of 0.0668

0.0668 = 6.68% probability that an individual man’s step length is less than 1.9 feet.

3 0

3 years ago

To plan the budget for next year a college must update its estimate of the proportion of next year's freshmen class that will ne

Naily [24]

Answer:

Null hypothesis: $p\leq 0.35$

Alternative hypothesis: $p > 0.35$

$z=\frac{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.491$

$p_v =P(z>2.491)=0.0064$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of applicants who request financial aid is significantly higher than 0.35

Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=67 represent the applicants who request financial aid

$\hat p=\frac{67}{150}=0.447$ estimated proportion of applicants who request financial aid

$p_o=0.35$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of applicants who request financial aid is higher than 0.35.:

Null hypothesis: $p\leq 0.35$

Alternative hypothesis: $p > 0.35$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.491$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>2.491)=0.0064$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of applicants who request financial aid is significantly higher than 0.35

5 0

3 years ago

I wasn’t sure if we were to bring ____ to the party.

Effectus [21]

Answer:

Presents

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Ella completed the following work to test the equivalence of two express

murzikaleks [220]

Answer:

The expressions are not equivalent because Ella did not know that you can’t use substitution to test for equivalence.

Step-by-step explanation:

<u>Equivalent algebraic expressions</u> are those expressions which on simplification give the same resulting expression.

Two algebraic expressions are said to be <u>equivalent</u> if their values obtained by substituting any values of the variables are same.

Two expressions 3f+2.6 and 2f+2.6 are not equivalent, because when f=1,

$3f+2.6=3\cdot 1+2.6=3+2.6=5.6\\ \\2f+2.6=2\cdot 1+2.6=2+2.6=4.6\\ \\5.6\neq 4.6$

Method of substitution can only help her to decide the expresssions are not equivalent, but if she wants to prove the expressions are equivalent, she must prove it for all values of f.

$3f+2.6=2f+2.6\\ \\3f=2f\\ \\3f-2f=0\\ \\f=0$

This is true only when f=0.

Hence,

The expressions are not equivalent because Ella did not know that you can’t use substitution to test for equivalence.

5 0

3 years ago

Read 2 more answers

A bag contains 8 red balls and 4 white balls. Two balls are drawn without replacement. (Enter your probabilities as fractions.)

RoseWind [281]

Answer:

total marbles;12

after drawing red total counts are 7 red and 4 white 11 in total

probabilities are 4/11

6 0

3 years ago