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Anna007 [38]
2 years ago
9

g A carload of steel rods has arrived at Cybermatic Construction Company. The car contains 50,000 rods. Claude Ong, manager of Q

uality Assurance, directs his crew measure the lengths of 100 randomly selected rods. If the population of rods has a mean length of 120 inches and a standard deviation of 0.05 inch, the probability that Claude's sample has a mean greater than 120.0125 inches is _____________.
Mathematics
1 answer:
Margarita [4]2 years ago
5 0

Answer:

0.0062

Step-by-step explanation:

Given that a carload of steel roads has arrived at Cybermatic Construction Company. The car contains 50,000 rods

Sample size =100

\bar x=120"

s =0.05"

Sample mean follows a N(120,0.05/sqrtn)

sample std error=\frac{0.05}{\sqrt{100} } \\=0.005

P(\bar X>120.0125)=\\P(Z>\frac{120.0125-120}{0.005} \\=P(Z>2.5)\\=0.5-0.4938\\=0.0062

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Answer:

$464

Step-by-step explanation:

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2 years ago
1) Consider the sequence:
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Answer:

1) The 38th term is A. 38

2) The 11th term is D. 3072

3) The 3rd term is C. 34

Step-by-step explanation:

1) I added 2 until I got to the 38th term.

2) I multiplied until I got to the 11th term.

3) I added 3 until i got to the 12th term.

The teacher showed a more complex way to do it but this is just what I did.

I took the quiz, so I know that I got them all correct.

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2 years ago
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What is the percentage decrease, if the price of a shirt is reduced from 800rupees to 640rupees?
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Decrease = Rs 800- Rs 640= Rs 160
Percentage decrease = (Decreased amount / Original amount) * 100
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2 years ago
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Assume the blood pressures of 10 people were measured before and after sleeping for the night. What is the corresponding critica
NeX [460]

Answer:

For the critical value we need to calculate the degrees of freedom given by:

df = n-1= 10-1=9

And since we have a one tailed test we need to look in the t distribution with 9 degrees of freedom a quantile who accumulates 0.05 of the area on a tail and we got:

|t_{crit}| =1.833

Step-by-step explanation:

Previous concepts

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.  

Let put some notation  

x=test value with right arm , y = test value with left arm  

The system of hypothesis for this case are:  

Null hypothesis: \mu_y- \mu_x = 0  

Alternative hypothesis: \mu_y -\mu_x \neq 0  

The first step is calculate the difference d_i=y_i-x_i

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{361}{5}  

The third step would be calculate the standard deviation for the differences, and we got:  

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}  

The 4 step is calculate the statistic given by :  

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}  

For the critical value we need to calculate the degrees of freedom given by:

df = n-1= 10-1=9

And since we have a one tailed test we need to look in the t distribution with 9 degrees of freedom a quantile who accumulates 0.05 of the area on a tail and we got:

|t_{crit}| =1.833

5 0
2 years ago
A marketing representative wants to estimate the proportion of people in a state who like the new design on the packaging of a c
Effectus [21]

Answer:

The correct option is;

(B) Yes, because sampling distributions of population proportions are modeled with a normal model.

Step-by-step explanation:

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2 years ago
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