Test 4 for AP Stats.
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Question 1 |
Each day, a teacher draws a student at random to be the line leader for the day. The teacher wants to see how many days on average it takes to get through each student out of a group of ten. Using as many trials as are possible in the random number table below, what is the probability it will take fewer than 20 days to get through all 10 students?
10442 76626 03118 42591 26005 54079 10968 88834
76724 74572 69273 2249 44581 30144 82816 30948
96774 08393 80202 27427 95995 16496 53676 71167
23236 77863 21731 20600 01139 54855 97484 72952
0 | |
0.17 | |
0.20 | |
0.25 |
Question 1 Explanation:
When using the table of random numbers, assign each student to a digit 0-9, and count the number of days that it takes to get through all 9 digits. The results are listed below.
10442 76626 03118 4259 (19 days)
1 26005 54079 10968 8883 (20 days)
4 76724 74572 69273 02249 44581 (26 days)
30144 82816 30948 96774 08393 80202 27427 95 (37 days)
995 16496 53676 71167 23236 77863 21731 20 (35 days)
600 01139 54855 97484 72 (20 days)
952 (remaining numbers not used)
The random number table produced 6 samples, of which 1 was less than 20 days (19). $1/6 ≈ 0.17$, so B is the correct answer.
10442 76626 03118 4259 (19 days)
1 26005 54079 10968 8883 (20 days)
4 76724 74572 69273 02249 44581 (26 days)
30144 82816 30948 96774 08393 80202 27427 95 (37 days)
995 16496 53676 71167 23236 77863 21731 20 (35 days)
600 01139 54855 97484 72 (20 days)
952 (remaining numbers not used)
The random number table produced 6 samples, of which 1 was less than 20 days (19). $1/6 ≈ 0.17$, so B is the correct answer.
Question 2 |
A student enters a drawing at a basketball game. There are 6 jerseys, 2 basketballs, and 20 free ticket packs. If 250 people enter the drawing, what is the probability the student will not win any prize?
1 | |
0.112 | |
0.08 | |
0.888 |
Question 2 Explanation:
The student has a 0.112 chance of winning $(6+2+20)÷250$. Therefore the probability of not winning is $1-0.112=0.888$ and answer D is correct.
Question 3 |
Which of the following are mutually exclusive events?
Being from Florida and knowing how to ice skate, because no one from Florida knows how to ice skate. | |
Rolling an odd number on a die and rolling a number less than three, because there is no overlap. | |
A flipped coin landing on heads and a flipped coin landing on tails, because these events cannot both occur on the same trial. | |
Drawing a heart from a deck of cards and drawing a face card, because these are independent events. |
Question 3 Explanation:
Answer C is correct. These two events cannot occur simultaneously; therefore, they are mutually exclusive.
Question 4 |
In a certain middle school, 28% of students play soccer and basketball, and 40% of students play basketball. What is the probability that a student plays soccer, given that the student plays basketball?
0.112 | |
0.12 | |
0.28 | |
0.7 |
Question 4 Explanation:
The formula for conditional probability is $\dfrac{P(AandB}{P(A)}$ where $P(A)$ is the probability of the first event and $P(B)$ is the probability of the second. $\dfrac{0.28}{0.40} = 0.7$ and answer choice D is correct.
Question 5 |
A student surveys students in the school to determine the rate of food allergies. The student finds that, out of a sample of 200 students, 5 have a peanut allergy, 2 have a nut allergy, 1 has a shellfish allergy, and 4 have an egg allergy. What is the probability that a student has either a peanut or nut allergy?
0.00025 | |
0.01 | |
0.025 | |
0.03475 |
Question 5 Explanation:
The probability of the union of events (or) is calculated by adding the probability of each event and subtracted the probability of both events occurring at once. In this case, $0.025+0.01-(0.025*0.01) = 0.03475$, and answer D is correct.
Question 6 |
A researcher surveyed 200 adults. Based on the results, the probability of a person taking a given number of pizza slices is displayed in the table below.
Number of slices |
Probability |
1 |
0.450 |
2 |
0.405 |
3 |
0.130 |
4 |
0.010 |
5 |
0.005 |
Based on this data, what is the average number of slices taken by a single individual?
1.715 | |
1.805 | |
1.940 | |
2.015 |
Question 6 Explanation:
The average number of languages can be found by $1 × .45 + 2 × .405 + 3 × .130 + 4 × .01 + 5 × .005$
$ = 1.715$. Answer choice A is correct.
$ = 1.715$. Answer choice A is correct.
Question 7 |
The probability distribution table for a discrete random variable is shown below
What are the mode, mean, and standard deviation of the data?
A. Mode = 3, Mean = 2.4, Standard Deviation ≈1.68 |
Question 7 Explanation:
The mode is the most frequently occurring number, in this case 3.
The mean is calculated by $μ_{x} = Σx_1*P(x_1)$
$\dfrac{1}{10} * 1 + \dfrac{2}{10} * 2 + \dfrac{3}{10} * 3 + \dfrac{2}{10} * 4 + \dfrac{1}{10} * 5 + \dfrac{1}{10} * 6$
$= 2.4 $
The standard deviation can be found by the formula $σ_x=\sqrt{Σ(x_1-μ_x )^2-P(x_1)}$
$\sqrt {\dfrac{1}{10} (1-2.4)^2 + \dfrac{1}{10} (2-2.4)^2 + \dfrac{3}{10} (3-2.4)^2 + \dfrac{2}{10} (4-2.4)^2 + \dfrac{1}{10} (5-2.4)^2 + \dfrac{1}{10} (6-2.4)^2 } ≈1.68$
The mean is calculated by $μ_{x} = Σx_1*P(x_1)$
$\dfrac{1}{10} * 1 + \dfrac{2}{10} * 2 + \dfrac{3}{10} * 3 + \dfrac{2}{10} * 4 + \dfrac{1}{10} * 5 + \dfrac{1}{10} * 6$
$= 2.4 $
The standard deviation can be found by the formula $σ_x=\sqrt{Σ(x_1-μ_x )^2-P(x_1)}$
$\sqrt {\dfrac{1}{10} (1-2.4)^2 + \dfrac{1}{10} (2-2.4)^2 + \dfrac{3}{10} (3-2.4)^2 + \dfrac{2}{10} (4-2.4)^2 + \dfrac{1}{10} (5-2.4)^2 + \dfrac{1}{10} (6-2.4)^2 } ≈1.68$
Question 8 |
At a craft fair, the average purchase amount is $30 with a standard deviation of $2. In addition, there is 7% sales tax plus a $5 charge to get in to the fair. Which of the following statements are true?
The mean and standard deviation of purchases with tax and entry are \$37.10 and \$2.14 | |
The mean and standard deviation of purchases with tax and entry are \$35 and \$2. | |
The mean and standard deviation of purchases with tax and entry are \$37.10 and \$2. | |
The mean and standard deviation of purchases with tax and entry are \$32.10 and \$0.14. |
Question 8 Explanation:
The transformed equation for this situation is $1.07X+5$. Therefore the mean is $1.07*30+5=\$37.10$ and the standard deviation is $1.07*2 = \$2.14$. Answer choice A is correct.
Question 9 |
A student rolls a die 36 times, with a success defined as rolling a 5 or 6. What are the mean and standard deviation for the binomial distribution?
Mean = 3.5, Standard Deviation = 1 | |
Mean = 12, Standard Deviation ≈2.828 | |
Mean = 3.5, Standard Deviation ≈ 2.828 | |
Mean = 12, Standard Deviation = 1 |
Question 9 Explanation:
For a binomial distribution, the mean is np where n = number of trials and p = probability of success. $36 * \dfrac{1}{3} = 12.$ The standard deviation is $\sqrt{np(1-p).}$ $\sqrt{36 * \dfrac{1}{3} (1 - \dfrac{1}{3})} = \sqrt{8} ≈ 2.828.$Answer choice B is correct.
Question 10 |
An automated dialer is dialing random numbers until it reaches a valid number. The mean for this situation is 12 and the standard deviation is approximately 11.5. Which statement is true?
The expected number of attempts before a valid number is reached is 11.5. | |
There will be approximately 0.5 valid numbers for every 12 attempts. | |
The expected number of attempts before a valid number is reached is 12. | |
The expected number of attempts before two valid numbers are reached is 23.5. |
Question 10 Explanation:
In a geometric distribution, the mean is the expected number of attempts before a success. Answer choice C is correct.
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