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ollegr [7]
1 year ago
10

Among 21- to 25-year-olds, 29% say they have driven while under the influence of alcohol. Suppose that three 21- to 25-year-olds

are selected at random.
a. What is the probability that at least 1 has driven while under the influence of alcohol?
b. Among 21- to 25-year-olds, 29% say they have driven while under the influence of alcohol.
Mathematics
1 answer:
Vilka [71]1 year ago
4 0

Answer:

0.6421

Step-by-step explanation:

In this case we have 3 trials and we have 2 options for each one. The driver has or hasn't been under alcohol influence. The probability that the driver has is 0.29 and the probabiility that the driver hasn't is 1 - 0.29 = 0.71

each trial is independent because we are assuming that the population of drivers in between 21 and 25 years old is very big.

The probability that one of them was under alcohol influence can be found by finding the probability that non of them was under alcohol influence because:

1 = p(x = 0) + p(x ≥ 1)

p(x ≥ 1) = 1 - p(0)

The probability that none of them was under alcohol influence is going to be:

0.71×0.71×0.71 = 0.3579

The probability of finding at least one driver that has been under alcohol influence is:

0.6421

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The table of this question is in the attachments.

Answer and Step-by-step explanation: <u>Standard</u> <u>deviation</u> is a measure of how spread the data is from the mean. It is calculated as:

s = √∑(x - μ)² / n - 1

where μ is the mean of the set.

<u>Range</u> is the difference between the highest and lowest value of a data set.

(a) <u>Range of Traditional course</u>:

range = 80.4 - 56

range = 24.4

<u>Range of "flipped" course</u>:

range = 92.6 - 63.5

range = 29.1

Comparing ranges, the "flipped" course has more dispersion than the traditional.

(b) <u>Standard Deviation of Traditional course</u>:

mean = 71.6

s = \sqrt{\frac{(70.3 - 71.6)^{2}+...+(59.1-71.6)^{2}}{13-1}

s = 8.95

<u>Standard Deviation of "flipped" course</u>:

mean = 77.6

s = \sqrt{\frac{(75.5 - 77.6)^{2}+...+(76.9-77.6)^{2}}{13-1}

s = 8.3

Comparing standard deviation, traditional course has more dispersion.

(c) If you change one score, range for traditional will be:

range = 591 - 56

range = 535

Changing one score increase in almost 22 times the range for this category.

7 0
1 year ago
The half life of a certain tranquilizer in the bloodstream is 37 hours. How long will it take for the drug to decay to 86% of th
nata0808 [166]

Answer:

  8.1 hours

Step-by-step explanation:

A model of the fraction remaining can be ...

  f = (1/2)^(t/37) . . . . t in hours

So, for the fraction remaining being 86%, we can solve for t using ...

  0.86 = 0.5^(t/37)

  log(0.86) = (t/37)log(0.5)

  t = 37·log(0.86)/log(0.5) ≈ 8.0509 ≈ 8.1 . . . hours

It takes about 8.1 hours to decay to 86% of the original concentration.

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1 year ago
A plane intersects the center of a sphere with a volume of about 113.1 m3. What is the area of the cross section? Round to the n
ivolga24 [154]

Answer:

Step-by-step explanation:

if ~radius=r \\volume =\frac{4}{3} \pi r^3\\113.1=\frac{4}{3} \pi r^3\\113.1 \times \frac{3}{4 \pi } =r^3\\\frac{339.3 }{4 \times 3.14} =r^3\\r^3=\frac{339.3}{12.56} \approx 27\\r=3\\area =\pi r^2=\pi *3^2=9\pi =9*3.14=28.26 \approx 28.3 ~m^2

5 0
2 years ago
For questions 2-5, the number of pieces in a regular bag of Skittles is approximately normally distributed with a mean of 38.4 a
aleksley [76]

Answer:

a. -1.60377

b. 0.25451

c. 0.344

d. Option b) 78th

Step-by-step explanation:

The number of pieces in a regular bag of Skittles is approximately normally distributed with a mean of 38.4 and a standard deviation of 2.12.

a)What is the z-score value of a randomly selected bag of Skittles that has 35 Skittles? a) 1.62 b) -1.62 c) 3.40 d) -3.40 e)1.303.

The formula for calculating a z-score is is z = (x-μ)/σ,

where x is the raw score

μ is the population mean

σ is the population standard deviation.

z = 35 - 38.4/2.12

= -1.60377

Option b) -1.62 is correct

b) What is the probability that a randomly selected bag of Skittles has at least 37 Skittles? a) .152 b) .247 c) .253 d).747e).7534. .

z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

z = (37 - 38.4)/2.12

= -0.66038

P-value from Z-Table:

P(x<37) = 0.25451

The probability that a randomly selected bag of Skittles has at least 37 Skittles is 0.25451

Option c) .253 is.correct

c) What is the probability that a randomly selected bag of Skittles has between 39 and 42 Skittles? a) .112 b) .232 c) .344 d).457 e).6125.

z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

For 39 Skittles

z = (39 - 38.4)/2.12

= 0.28302

Probability value from Z-Table:

P(x = 39) = 0.61142

For 42 Skittles

z = (42 - 38.4)/2.12

= 1.69811

Probability value from Z-Table:

P(x = 42) = 0.95526

The probability that a randomly selected bag of Skittles has between 39 and 42 Skittles is:

P(x = 42) - P(x = 39

0.95526 - 0.61142

0.34384

= 0.344

Option c is.correct

d) What is the percentile rank of a randomly selected bag of Skittles that has 40 Skittles in it? a)82nd b) 78th c) 75th d)25th e)22nd

z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

z = (40 - 38.4)/2.12

= 0.75472

P-value from Z-Table:

P(x = 40) = 0.77479

Converting to percentage = 0.77479× 100

= 77. 479%

≈ 77.5

Percentile rank = 78th

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Answer:

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