Ask question

Ask question

All categories

3 years ago

9

A bank's loan officer rates applicants using their FICO credit score. The ratings are normally distributed with a mean of 711 an

d a standard deviation of 80. Find the credit score which separates the lower 80% from the top 20%. g

1 answer:

tekilochka [14]3 years ago

6 0

Answer:

The answer is "778"

Step-by-step explanation:

Given value:

$\to \mu= 711\\\\\to \sigma= 80\\\\\to P(Z >0.84) = 0.20\\\\\to z = 0.84$

The formula for z-score:

$\to x = z \times \sigma + \mu \\\\\to x = 0.84 \times 80 + 711 \\\\ \to x=67.2 +711\\\\ \to x= 778.2\\\\ \to x= 778$

You might be interested in

Morgan scored 41 points in 5 games. How many points would you expect him to make in an 21 game season. Round your answer to the

faust18 [17]

Answer: 172

Step-by-step explanation:

If what I did is right, the answer is 172.2 since 41/5 is 8.2 and I multiply 8.2 times 21, I get 172.2, which rounds to 172.

3 0

3 years ago

According to a Washington Post-ABC News poll, 331 of 502 randomly selected U.S. adults interviewed said they would not be bother

k0ka [10]

Answer:

$z=\frac{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{502}}}=7.124$

$p_v =P(z>7.124)=5.24x10^{-13}$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ 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 interest is significantly higher than 0.5.

Step-by-step explanation:

Data given and notation

n=502 represent the random sample taken

X=331 represent the adults that said they would not be bothered if the NAtional security agency

$\hat p=\frac{331}{502}=0.659$ estimated proportion of people who would not be bothered if the NAtional security agency

$p_o=0.5$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

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 is higher than the majority of 0.5:

Null hypothesis: $p\leq 0.5$

Alternative hypothesis: $p > 0.5$

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:

$z=\frac{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{502}}}=7.124$

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 significance level provided $\alpha=0.01$ . The next step would be calculate the p value for this test.

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

$p_v =P(z>7.124)=5.24x10^{-13}$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ 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 interest is significantly higher than 0.5.

4 0

3 years ago

What is a rational number?

MAVERICK [17]

In mathematics, a rational number is a number that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non-zero denominator q. Since q may be equal to 1, every integer is a rational number.

4 0

3 years ago

Read 2 more answers

1) On a standardized aptitude test, scores are normally distributed with a mean of 100 and a standard deviation of 10. Find the

Musya8 [376]

Answer:

A) 34.13%

B) 15.87%

C) 95.44%

D) 97.72%

E) 49.87%

F) 0.13%

Step-by-step explanation:

To find the percent of scores that are between 90 and 100, we need to standardize 90 and 100 using the following equation:

$z=\frac{x-m}{s}$

Where m is the mean and s is the standard deviation. Then, 90 and 100 are equal to:

$z=\frac{90-100}{10}=-1\\ z=\frac{100-100}{10}=0$

So, the percent of scores that are between 90 and 100 can be calculated using the normal standard table as:

P( 90 < x < 100) = P(-1 < z < 0) = P(z < 0) - P(z < -1)

= 0.5 - 0.1587 = 0.3413

It means that the PERCENT of scores that are between 90 and 100 is 34.13%

At the same way, we can calculated the percentages of B, C, D, E and F as:

B) Over 110

$P( x > 110 ) = P( z>\frac{110-100}{10})=P(z>1) = 0.1587$

C) Between 80 and 120

$P( 80$

D) less than 80

$P( x < 80 ) = P( z$

E) Between 70 and 100

$P( 70$

F) More than 130

$P( x > 130 ) = P( z>\frac{130-100}{10})=P(z>3) = 0.0013$

8 0

3 years ago

What five-letter word becomes shorter when you add two letters to it?

Fantom [35]

you literally just need to add a single letter. You need to place the letters “t” and “r” to the word “shore”. After you do that, the word becomes “shorter”. Another possible answer is the word “Sorte”.

7 0

3 years ago

Read 2 more answers