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2 years ago

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Q. If heights of 3rd graders follow a normal distribution with a mean of 52 inches and a standard deviation of 2.5 inches, what

is the z score of a 3rd grader who is 47 inches tall?
A. z=-5
B. z=-2
C. z=2
D. z=5

1 answer:

zaharov [31]2 years ago

3 0

Answer:

-2

Step-by-step explanation:

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It has been suggested that night shift-workers show more variability in their output levels than day workers. Below, you are giv

bonufazy [111]

Answer:

Null hypotheses = H₀ = σ₁² ≤ σ₂²

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic = 1.9

p-value = 0.206

Since the p-value is greater than α therefore, we cannot reject the null hypothesis.

So we can conclude that the night shift workers don't show more variability in their output levels than day workers.

Step-by-step explanation:

Let σ₁² denotes the variance of night shift-workers

Let σ₂² denotes the variance of day shift-workers

State the null and alternative hypotheses:

The null hypothesis assumes that the variance of night shift-workers is equal to or less than day-shift workers.

Null hypotheses = H₀ = σ₁² ≤ σ₂²

The alternate hypothesis assumes that the variance of night shift-workers is more than day-shift workers.

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic:

The test statistic or also called F-value is calculated using

Test statistic = Larger sample variance/Smaller sample variance

The larger sample variance is σ₁² = 38

The smaller sample variance is σ₂² = 20

Test statistic = σ₁²/σ₂²

Test statistic = 38/20

Test statistic = 1.9

p-value:

The degree of freedom corresponding to night shift workers is given by

df₁ = n - 1

df₁ = 9 - 1

df₁ = 8

The degree of freedom corresponding to day shift workers is given by

df₂ = n - 1

df₂ = 8 - 1

df₂ = 7

We can find out the p-value using F-table or by using Excel.

Using Excel to find out the p-value,

p-value = FDIST(F-value, df₁, df₂)

p-value = FDIST(1.9, 8, 7)

p-value = 0.206

Conclusion:

p-value > α

0.206 > 0.05 ( α = 1 - 0.95 = 0.05)

Since the p-value is greater than α therefore, we cannot reject the null hypothesis corresponding to a confidence level of 95%

So we can conclude that the night shift workers don't show more variability in their output levels than day workers.

3 0

2 years ago

Dan is using tiles to add Negative 8 + 6. He begins with the tiles shown below. 8 negative tiles. What is the sum of Negative 8

navik [9.2K]

The answer is -2

-8+6=-2

4 0

2 years ago

Read 2 more answers

How can we find how many inches wide the coat box is?

Alex777 [14]

Answer:

First we found the sum of 15.438 and 3.18. The subtract the answer from 32.476.

Option A is correct.

Step-by-step explanation:

The total width of box is = 32.476

Width of box of T-shirts = 15.438

Width of box of Socks = 3.18

We need to find width of coat box.

The width of coat box can be found as:

Total Width = Width of box of T-shirts + Width of box of Socks + width of coat box

32.476 = 15.438 + 3.18 + width of coat box

32.476 = 18.618 + width of coat box

width of coat box= 32.476 - 18.618

width of coat box = 13.858

So, The Width of coat box = 13.858 inches

First we found the sum of 15.438 and 3.18. The subtract the answer from 32.476.

Option A is correct.

5 0

1 year ago

comparing permutations to combinations for the same set of parameters you would have more combinations than permutations true or

ella [17]

Let a set of $n$ elements.
We can find $n!$ (factorial) of the $n$ element.
However, combination of the element lead to less than $n!$ possibilities.
(combining like adding or multiplying)
So the proposition is false.

6 0

2 years ago

Read 2 more answers

In the isosceles △ABC m∠ACB=120° and AD is an altitude to leg BC . What is the distance from D to base AB , if CD=4cm?

7nadin3 [17]

Correct answer is: distance from D to AB is 6cm

Solution:-

Let us assume E is the altitude drawn from D to AB.

Given that m∠ACB=120° and ABC is isosceles which means

m∠ABC=m∠BAC = $\frac{180-120}{2}=30$

And AC= BC

Let AC=BC=x

Then from ΔACD , cos(∠ACD) = $\frac{DC}{AC} =\frac{4}{x}$

Since DCB is a straight line m∠ACD+m∠ACB =180

m∠ACD = 180-m∠ACB = 60

Hence $cos(60)=\frac{4}{x}$

$x=\frac{4}{cos60}= 8$

Now let us consider ΔBDE, sin(∠DBE) = $\frac{DE}{DB} =\frac{DE}{DA+AB} = \frac{DE}{4+8}$

$DE = 12sin(30) = 6cm$

7 0

2 years ago