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2 years ago

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The scores of eighth-grade students in a math test are normally distributed with a mean of 57.5 and a standard deviation of 6.5.

From this data, we can conclude that 68% of the students received scores between and .

2 answers:

Anna11 [10]2 years ago

8 0

Approximately 68% of a normal distribution lies within one standard deviation of the mean, so this corresponds to students with scores between (57.5 - 6.5, 57.5 + 6.5) = (51, 64)

Kaylis [27]2 years ago

5 0

As given, the scores of eighth-grade students in a math test are normally distributed with a mean of 57.5 and a standard deviation of 6.5.

This means that around 68% of the distributed data lies within the ranges of -6.5 to +6.5 of the given standard deviation.

This implies to:

$57.5-6.5=51$ and $57.5+6.5=64$

So, the range of scores lies between (51,64)

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SAT Writing scores are normally distributed with a mean of 491 and a standard deviation of 113.A university plans to send letter

Sholpan [36]

Answer:

$z=1.405$

And if we solve for a we got

$a=491 +1.405*113=649.765$

So the value of height that separates the bottom 92% of data from the top 8% is 649.765.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

$X \sim N(491,113)$

Where $\mu=491$ and $\sigma=113$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.08$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.92 of the area on the left and 0.08 of the area on the right it's z=1.405. On this case P(Z<1.405)=0.92 and P(z>0.92)=0.08

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.405$

And if we solve for a we got

$a=491 +1.405*113=649.765$

So the value of height that separates the bottom 92% of data from the top 8% is 649.765.

6 0

2 years ago

Please help!!!!!!!!!

shepuryov [24]

A, the first one only, this parabola only has a minimum and no maximum. the other statements are also just false

7 0

2 years ago

A sample of 75 concrete blocks had a mean mass of 38.3 kg with a standard deviation of 0.6 kg.

Scilla [17]

The answer will be a

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2 years ago

A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win $1.10; if

topjm [15]

Answer:

a) The expected value is $\frac{-1}{15}$

b) The variance is $\frac{49}{45}$

Step-by-step explanation:

We can assume that both marbles are withdrawn at the same time. We will define the probability as follows

#events of interest/total number of events.

We have 10 marbles in total. The number of different ways in which we can withdrawn 2 marbles out of 10 is $\binom{10}{2}$ .

Consider the case in which we choose two of the same color. That is, out of 5, we pick 2. The different ways of choosing 2 out of 5 is $\binom{5}{2}$ . Since we have 2 colors, we can either choose 2 of them blue or 2 of the red, so the total number of ways of choosing is just the double.

Consider the case in which we choose one of each color. Then, out of 5 we pick 1. So, the total number of ways in which we pick 1 of each color is $\binom{5}{1}\cdot \binom{5}{1}$ . So, we define the following probabilities.

Probability of winning: $\frac{2\binom{5}{2}}{\binom{10}{2}}= \frac{4}{9}$

Probability of losing $\frac{(\binom{5}{1})^2}{\binom{10}{2}}\frac{5}{9}$

Let X be the expected value of the amount you can win. Then,

E(X) = 1.10*probability of winning - 1 probability of losing = $1.10\cdot \frac{4}{9}-\frac{5}{9}=\frac{-1}{15}$

Consider the expected value of the square of the amount you can win, Then

E(X^2) = (1.10^2)*probability of winning + probability of losing = $1.10^2\cdot \frac{4}{9}+\frac{5}{9}=\frac{82}{75}$

We will use the following formula

$Var(X) = E(X^2)-E(X)^2$

Thus

Var(X) = $\frac{82}{75}-(\frac{-1}{15})^2 = \frac{49}{45}$

7 0

2 years ago

Riya is applying mulch to her garden. She applies it at a rate of 250{,}000\text{ cm}^3250,000 cm 3 250, comma, 000, start text,

koban [17]

Complete Question

Riya is applying to her garden. She applies it at a rate of 25, 000 cm³ of mulch for every m² of garden space. At what rate is Riya applying mulch in m³/m²

Answer:

0.25m³/m²

Step-by-step explanation:

We are told the Riya sprays mulch at 250,000cm³ per m²

To find the rate at which Riya is spaying the mulch in m³/m² we would have to convert 250,00cm³/m² to m³/m²

1 cm³ = 1 × 10^-6m³

250,000 cm³ = x m³

Cross Multiply

1 cm³ × xm³ = 250,000cm³ × 1 × 10^-6 m³

X

x m³ = 250,000cm³ × 1 × 10^-6 m³/1 cm³

= 0.25m³

Therefore, the rate at which Riya is spraying the mulch = 0.25m³/m²

3 0

1 year ago