Which is better, a score of 92 on a test with a mean of 71 and a standard

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 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:

brainly.com/question/14777840

Solution 2

Guest 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

So

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

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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.

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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:

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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

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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 .

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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
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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