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Which is better, a score of 92 on a test with a mean of 71 and a standard deviation of 15, or a score of 688 on a test with a mean of 493 and a standard deviation of 150? a. Both scores have the same relative position b. A score of 92 c.A score of 688

2 months ago

Solution 1

Guest #2392415

2 months ago

The z-score for a score of 92 is higher than the z-score of a score of 688, therefore, the score that is better is: b. A score of 92

Recall:

In comparing scores or determining how relatively far scores are from the mean in a distribution, we can transform the score using the z-score.

Z-score = (raw score - mean)/standard deviation.

Z-score for a score of 92:

raw score = 92

mean = 71

standard deviation = 15

Z-score = (92 - 71)/15 = 1.4

Z-score for a score of 688:

raw score = 688

mean = 493

standard deviation = 150

Z-score = (688 - 493)/150 = 1.3

Therefore, the z-score for a score of 92 is higher than the z-score of a score of 688, therefore, the score that is better is: b. A score of 92

Learn more about z-score on:

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Solution 2

Guest #2392416

2 months ago

Answer:

b. A score of 92

Step-by-step explanation:

The z-score measures how many standard deviations a score is above or below the mean.

It is given by the following formula:

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

In which $X$ is the score, $\mu$ is the mean and $\sigma$ is the standard deviation.

In this problem, we have that:

The best score is the one with a higher z-score. If the z-score is the same for both, then they have the same relative position.

A score of 92 on a test with a mean of 71 and a standard deviation of 15.

Here we have $X = 92, \mu = 71, \sigma = 15$

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

$Z = \frac{92 - 71}{15}$

$Z = 1.4$

A score of 688 on a test with a mean of 493 and a standard deviation of 150

Here we have $X = 688, \mu = 493, \sigma = 150$

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

$Z = \frac{688 - 493}{150}$

$Z = 1.3$

The score of 92 has a higher Z-score, so it is better.

The correct answer is:

b. A score of 92

📚 Related Questions

Question

An aptitude test has a mean score of 80 and a standard deviation of 5. The population of scores is normally distributed. What proportion of tests has scores over 90?

Solution 1

Answer:

2.28% of tests has scores over 90.

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:

$\mu = 80, \sigma = 5$

What proportion of tests has scores over 90?

This proportion is 1 subtracted by the pvalue of Z when X = 90. So

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

$Z = \frac{90 - 80}{5}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772.

So 1-0.9772 = 0.0228 = 2.28% of tests has scores over 90.

Question

The maturity value of a 50-day loan of $550 is $560. What is the annual simple interest rate (in percent) on this loan? Round to the nearest tenth of a percent. Use 360 days in 1 year.

Solution 1

Answer:

13.1%

Step-by-step explanation:

Principal value of the loan = $550

Maturity Value of the loan = $560

Let the rate of simple interest = R

Time = 50 days = $\frac{50}{360}$ years

Simple Interest = Maturity Value - Principal Value = $560 - $550 = $10

But Simple Interest = $\frac{Principal * R * Time}{100}$

Substituting, $10 = \frac{550 * R * 50/360}{100}$

$=> R = \frac{10*100*360}{550*50}$

$=> R = 13.1%$

Question

A delivery of 50 transistors contains 40 good ones and 10 defectives. In a test five of them are checked. How many possibilities are there to have 3 good ones and 2 defective transistors in the test set?

Solution 1

Answer:

0.2048

Step-by-step explanation:

p(good transistor) = 40/50 = 4/5

p(defective transistor) = 10/50 = 1/5

the probability of having 3 good ones and 2 defective transistors in the test set :

= 5C3×(4÷5)³×(1÷5)²

= 0.2048

Solution 2

Answer:

0.2048

Step-by-step explanation:

Question

You are standing next to a really big circular lake. You want to measure the diameter of the lake, but you don't want to have to swim across with a measuring tape! You decide to walk around the perimeter of the lake and measure its circumference, and find that it's 4007 m.

Solution 1

Answer:

1 275.5 m

Step-by-step explanation:

Measure of the circumference = 4007

on the other hand ,the Measure of the circumference = (diameter)×π

then

the measure of the diameter = 4 007÷π = 1 275.46771393845 m

Question

A certain delivery service offers both express and standard delivery. 75% of parcels are sent by standard delivery, and 25% are sent by express. Of those sent standard, 80% arrive the next day, and of those sent express, 95% arrive the next day. A record of parcel delivery is chosen at random from the company's files.What is the probability that the parcel was shipped expressed and arrived the next day?

Solution 1

Answer:

The probability that the parcel was shipped expressed and arrived the next day = 0.2375 .

Step-by-step explanation:

We are given that a certain delivery service offers both express and standard delivery.

Let Proportion of parcels sent by standard delivery, P( $A_1$ ) = 0.75

Proportion of parcels sent by express delivery, P( $A_2$ ) = 0.25

Let event B = Parcel arriving the next day

Also, Probability of parcel arriving the next day given it was sent through standard delivery, P(B/ $A_1$ ) = 0.8

Probability of parcel arriving the next day given it was sent through express delivery, P(B/ $A_2$ ) = 0.95

Now, Probability that the parcel was shipped expressed and arrived the next day = P( $A_2$ ) * P(B/ $A_2$ ) = 0.25 * 0.95 = 0.2375 .

Question

Suppose you are the manager of a firm. The accounting department has provided cost estimates, and the sales department sales estimates, on a new product. Analyze the data they give you, determine what it will take to break even, and decide whether to go ahead with production of the new product. The product has a production cost function C(x)equals75xplus3 comma 350 and a revenue function R(x)equals100x.

Solution 1

Answer:

If more than 134 articles are produced and sold, the firm will have positive profits and shoule start production

Step-by-step explanation:

Cost, Revenue, and Profit Function

The cost function C(x) is given by

$C(x)=75x+3,350$

where x is the number of produced products.

The revenue function is

$R(x)=100x$

With both equations, we can know the profit function as

$P(x)=R(x)-C(x)$

$P(x)=100x-75x-3,350=25x-3,350$

$P(x)=25x-3,350$

For the firm to have positive profits, it has to produce x articles with the condition

$P(x)>0$

Or, equivalently

$P(x)=25x-3,350>0$

Solving for x

$\displaystyle x>\frac{3,350}{25}$

Thus

$x>134$

This means that if more than 134 articles are produced and sold, the firm will have positive profits and shoule start production

Question

A test has 6 multiple choice questions, each with 4 alternatives. What is the probability of guessing 5 or more questions correctly?

Solution 1

Answer:

0.0009765625‬

Step-by-step explanation:

Each multiple choice question has 4 equally likely alternatives.

Hence, probability of guessing the right answer for 1 question is $\frac{1}{4}$ = 0.25

Hence, probability of guessing the wrong answer is $\frac{1}{4}$ = 0.75

The probability of guessing 5 or more questions correctly = probability of guessing 5 questions correctly & 1 question wrong + probability of guessing 6 questions correctly

= $\[0.25^{5}*0.75+0.25^{6}\]
$

= 0.000732421875 + 0.000244140625

= 0.0009765625‬

Question

Find the x,y and z intercepts of the plane tangent to the sphere of radius sqrt(14), with center at the origin, at the point (1,2,3)

Solution 1

The equation of the tangent of the plane will be equal to 2 (x - 1) +4 (y - 2) + 6(z - 3) = 0

What is a sphere?

A three-dimensional concrete figure with a spherical-like shape is known as a spherical in geometry. It is a collection of points in three dimensions that are linked by a single common point and are spaced equally apart.

As per the data given in the question,

f(x) = 2x

f(y) = 2y and,

f(z) = 2z

At points (1, 2, 3),

For $f_x$ (1, 2, 3) = 2(1) = 2

For $f_y$ (1, 2, 3) = 2(2) = 4

For $f_z$ (1, 2, 3) = 2(3) = 6

The equation for normal,

n = 2i + 4j + 6k

The equation for the tangent of the plane,

2 (x - 1) +4 (y - 2) + 6(z - 3) = 0

The x-intercept can be found when y and z are 0.

To know more about sphere:

brainly.com/question/11374994

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Solution 2

It would be (2,3) because of the 14 and 23

Question

A fair die is rolled 10 times. What is the probability that an odd number (1, 3, or 5) will occur fewer than 3 times?

Solution 1

Answer:

The probability that an odd number rolls of a die for less than 3 times is 0.054.

Step-by-step explanation:

The sample space of rolling a fair die is, S = {1, 2, 3, 4, 5, 6}

The odd numbers are, {1, 3, and 5}.

The probability that an odd number occurs is:

$P(Odd)=\frac{Favorable\ outcomes}{Total\ no.\ of\ outcomes}=\frac{3}{6}=\frac{1}{2}$

The die was rolled n = 10 times.

Let X = number of rolls in which an odd number occurs.

The random variable $X\sim Bin(n=10,p=\frac{1}{2})$

The probability distribution of binomial is:

$P(X =x)={n\choose x}p^{x}(1-p)^{n-x}$

Compute the probability that an odd number will occur less than 3 times as follows:

$P(X$

Thus, the probability that an odd number rolls of a die for less than 3 times is 0.054.

Question

"The weight of a product is normally distributed with a mean of four ounces and a variance of .25 squared ounces. What is the probability that a randomly selected unit from a recently manufactured batch weighs no more than 3.5 ounces?

Solution 1

Answer:

Step-by-step explanation:

Since the weight of the products are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = weight of the product.

µ = mean weight

σ = standard deviation

From the information given,

µ = 4 ounces

Variance = 0.25²

σ = √variance = √0.25² = 0.25

We want to find the probability that a randomly selected unit from a recently manufactured batch weighs no more than 3.5 ounces. It is expressed as

P(x ≤ 3.5)

For x = 3.5,

z = (3.5 - 4)/0.25 = - 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.0228

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