Number 9, A and B please

Among freshman at a certain university, scores on the Math SAT followed the normal curve, with an average of 550 and an SD of 10

lina2011 [118]

Answer:

a) 6.68th percentile

b) 617.5 points

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 550, \sigma = 100$

a) A student who scored 400 on the Math SAT was at the ______ th percentile of the score distribution.

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

$Z = \frac{400 - 550}{100}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

So this student is in the 6.68th percentile.

b) To be at the 75th percentile of the distribution, a student needed a score of about ______ points on the Math SAT.

He needs a score of X when Z has a pvalue of 0.75. So X when Z = 0.675.

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

$0.675 = \frac{X - 550}{100}$

$X - 550 = 0.675*100$

$X = 617.5$

6 0

3 years ago

Wind farm configuration is a significant issue as the upwind turbines generate maximal energy while creating wakes for downwind

lord [1]

Answer:

a) [24.114,29.2858]

b) Since 30 is not in the 95% confidence interval, there is a 95% probability that 30 is not the true mean WEP

Step-by-step explanation:

a)

The 95% confidence interval is given by the interval

$\bf [ \bar x-z^*\frac{s}{\sqrt n}, \bar x+t^*\frac{s}{\sqrt n}]$

where

$\bf \bar x$ = 26.7 is the sample mean

s = 17.7 is the sample standard deviation

n = 180 is the sample size

Since the sample size is big enough, we can use the Normal N(0,1) to compute $\bf z^*$ and it would be 1.96(*) (a value such that the area under the Normal curve outside the interval [-z, z] is 5% (0.05))

and our 95% confidence interval is

$\bf [26.7-1.96*\frac{17.7}{\sqrt{180}}, 26.7+1.96*\frac{17.7}{\sqrt{180}}]=\boxed{[24.114,29.2858]}$

(*)

This value can be computed in Excel with

<em>NORMINV(1-0.025,0,1)</em>

and in OpenOffice Calc with

<em>NORMINV(1-0.025;0;1)</em>

b)

Since 30 is not in the 95% confidence interval, there is a 95% probability that 30 is not the true mean WEP

7 0

3 years ago

The length of the hypotenuse of a right triangle is 14 inches. If the length of one leg is 6 inches, what is the approximate len

Liula [17]

Using the Pythagorean theorem:

6^2 + X^2 = 14^2

36 + x^2 = 196

X^2 = 196-36

x^2 = 160

x = SQRT(160)

x = 4 SQRT(10)
Or as a decimal:
x = 12.65 inches

5 0

3 years ago

Read 2 more answers

How Do I solve this I am quite confused

Kazeer [188]

Answer:

$11,714

Step-by-step explanation:

C(x) = 0.7x² - 462x + 87,944

this is a quadratic equation that opens up

the vertex will be the minimum

From the quadratic formula the x coordinate of the vertex is

x = -b/2a

x = 462/(2 * 0.7)

x = 330

Plug in

c(330) = 0.7(330²) - 462(330) + 87,944

c(330) = 11,714

$11,714

5 0

3 years ago

A phone company advertises a new plan in which the customer pays a fixed amount of $25 per month for unlimited calls in the coun

GarryVolchara [31]

Answer 750$ becaus jfhfhfhfhfhfhfhfhfhfh

8 0

2 years ago

Read 2 more answers