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victus00 [196]
3 years ago
14

On each trial of a digit span memory task, the participant is asked to read aloud a string of random digits. The participant mus

t then repeat the digits in the correct order. If the participant is successful, the length of the next string is increased by one. For instance, if the participant repeats four digits successfully, she will hear five random digits on the next trial. The participant’s score is the longest string of digits she can successfully repeat.
A professor of cognitive psychology is interested in the number of digits successfully repeated on the digit span task among college students. She measures the number of digits successfully repeated for 64 randomly selected students. The professor knows that the distribution of scores is normal, but she does not know that the true average number of digits successfully repeated on the digit span task among college students is 7.06 digits with a standard deviation of 1.610 digits.
The expected value of the mean of the 64 randomly selected students, M, is7.06 . (Hint: Use the population mean and/or standard deviation just given to calculate the expected value of M.)
The standard error of M is1.610 . (Hint: Use the population mean and/or standard deviation just given to calculate the standard error.)
The DataView tool that follows displays a data set consisting of 200 potential samples (each sample has 64 observations).
Suppose this professor happens to select Sample 103. (Hint: To see a particular sample, click the Observations button on the left-hand side of the DataView tool. The samples are numbered in the first column, and you can use the scroll bar on the right side to scroll to the sample you want.)
Use the DataView tool to find the mean and the standard deviation for Sample 103. The mean for Sample 103 is7.06 . The standard deviation for Sample 103 is .
Using the distribution of sample means, calculate the z-score corresponding to the mean of Sample 103. The z-score corresponding to the mean of Sample 103 is .
Use the Distributions tool that follows to determine the probability of obtaining a mean number of digits successfully repeated greater than the mean of Sample 103.
The probability of obtaining a sample mean greater than the mean of Sample 103 is .
If the sample you select for your statistical study is 1 of the 200 samples you drew in your repeated sampling, the worst-luck sample you could draw is . (Hint: The worst-luck sample is the sample whose mean is farthest from the true mean. You may find it helpful to sort the sample means: In Observations view click the arrow below the column heading Sample Means.)
Mathematics
1 answer:
Sphinxa [80]3 years ago
8 0
Does Warren support a bill rights
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yaroslaw [1]

Answer:

a) 59.34%

b) 44.82%

c) 26.37%

d) 4.19%

Step-by-step explanation:

(a)

There are in total <em>4+5+6 = 15 bulbs</em>. If we want to select 3 randomly there are  K ways of doing this, where K is the<em> combination of 15 elements taken 3 at a time </em>

K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455

As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.

So, the probability of selecting exactly 2 bulbs of 75 W is

\frac{270}{455}=0.5934=59.34\%

(b)

The probability of selecting three 40-W bulbs is

\frac{4*3*2}{455}=0.0527=5.27\%

The probability of selecting three 60-W bulbs is

\frac{5*4*3}{455}=0.1318=13.18\%

The probability of selecting three 75-W bulbs is

\frac{6*5*4}{455}=0.2637=26.37\%

Since <em>the events are disjoint</em>, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%

(c)

There are 6*5*4 ways of selecting one bulb of each type, so the probability of selecting 3 bulbs of each type is

\frac{6*5*4}{455}=0.2637=26.37\%

(d)

The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, <em>supposing there is no replacement</em>, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.

As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is \frac{9}{15}=0.6

Since there are no replacement, the probability of taking a second non 75-W bulb is now \frac{8}{14}=0.5714

Following this procedure 5 times, we find the probabilities

\frac{9}{15},\frac{8}{14},\frac{7}{13},\frac{6}{12},\frac{5}{11}

which are

0.6, 0.5714, 0.5384, 0.5, 0.4545

As the events are independent, the probability of choosing 5 non 75-W bulbs is the product

0.6*0.5714*0.5384*0.5*0.4545 = 0.0419 = 4.19%

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Step-by-step explanation:

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yanalaym [24]

Answer:

X = 5

Step-by-step explanation:

Here is what I am picturing (not to scale)

A ----------------B-----------C

We are given the information of AB (which equals 3x + 1), BC (which equals 4x - 5), and AC (which equals 8x - 9).

With reference to the line, AB + BC = AC

So,

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