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Delicious77 [7]

2 years ago

14

In a sample of 100 steel canisters, the mean wall thickness was 8.1 mm with a standard deviation of 0.5 mm. Someone says that th

e mean thickness is less than 8.2 mm. With what level of confidence can this statement be made? (Express the final answer as a percent and round to two decimal places.)

1 answer:

Travka [436]2 years ago

8 0

Answer:

This statement can be made with a level of confidence of 97.72%.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 8.1 mm

Standard Deviation, σ = 0.5 mm

Sample size, n = 100

We are given that the distribution of thickness is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling:

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.5}{\sqrt{100}} = 0.05$

P(mean thickness is less than 8.2 mm)

P(x < 8.2)

$P( x < 8.2)\\\\ = P( z < \displaystyle\frac{8.2 - 8.1}{0.05})\\\\ = P(z < 2)$

Calculation the value from standard normal z table, we have,

$P(x < 8.2) =0.9772 = 97.72\%$

This statement can be made with a level of confidence of 97.72%.

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Yuri [45]

Answer:

D. n1 and n2 can be of different sizes

Step-by-step explanation:

When we are trying to develop interval estimate for difference between two populations, we first select two samples from these populations. We get the interval using the required statistic.

These are:

- Sample size of both populations. Which are n1 and n2 respectively.

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4 0

2 years ago

troy's truck has a 30 gallon gas tank and gets an average of 21 miles per gallon.write an equation to represent the amount of ga

GuDViN [60]

Let

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y--------> represent the amount of gas in Troy's truck

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$y-y1=m*(x-x1)$

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in this problem the slope is equal to $-\frac{1}{21} \frac{gallons}{miles}$

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$for\ x=0\ y=30\ gal$

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8 0

2 years ago

Read 2 more answers

Adrian ran from his house to the park at a speed of 5 meters per second. He ran back from the park to his house at a speed of 6

denpristay [2]

Answer:time taken to run from his house to the park is 12 minutes

Step-by-step explanation:

Let x represent the distance between his house and the park.

The speed at which Adrian ran from his house to the park is 5 meters per second.

Distance = speed × time

Time = distance/speed

Therefore, time taken to run from his house to the park would be

Time = x/5

The speed at which he ran back from the park to his house is 6 meters per second.

Time taken to run back from the park to his house would be

Time = x/6

The total time taken to run the whole journey was 22 minutes. Converting to seconds, it becomes 22× 60 = 1320 seconds. Therefore,

x/5 + x/6 = 1320

Multiplying through by 30, it becomes

6x + 5x = 39600

11x = 39600

x = 39600/11 = 3600 meters

Therefore, time taken to run from his house to the park would be

x/5 = 3600/5 = 720 seconds

Converting to minutes, it becomes

720/60 = 12 minutes

7 0

2 years ago

shanti wrote the predicted values for a data set using the line of best fit y = 2.55x – 3.15. she computed two of the residual v

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2 years ago

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A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defe

xxMikexx [17]

Answer:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

Step-by-step explanation:

For the random variable $N_1$ we define the possible values for this variable on this case $[1,2,3,4,5]$ . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

$5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10$

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable $N_1$ like this:

$P(N_1 = a) = \frac{5-a C 1}{5C2}$

For the distribution of $N_2$ we need to take in count that we are finding a conditional distribution. $N_2$ given $N_1 =a$ , for this case we see that $N_2 \in [1,2,...,5-a]$ , so then exist $5-a C 1$ ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:

$P(N_2 =b | N_1 = a) = \frac{1}{5-a C 1}$

And if we want to find the joint probability we just need to do this:

$P(N_1 = a , N_2 = b) = P(N_2 = b | N_1 = a) P(N_1 =a)$

And if we multiply the probabilities founded we got:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

8 0

2 years ago