0.2 Practice tests (1-4) and final exams  (Page 11/36)

31 . The domain of X = {1, 2, 3, 4, 5, 6, 7, 8., 9, 10, 11, 12…27}. Because you are drawing without replacement, and 26 of the 52 cards are red, you have to draw a red card within the first 17 draws.

32 . X ~ G (0.24)

33 . $\mu =\frac{1}{p}=\frac{1}{0.27}=3.70$

34 . $\sigma =\sqrt{\frac{1-p}{p^2}}=\sqrt{\frac{1-0.27}{{0.27}^2}}=3.16$

4.5: hypergeometric distribution

35 . Yes, because you are sampling from a population composed of two groups (boys and girls), have a group of interest (boys), and are sampling without replacement (hence, the probabilities change with each pick, and you are not performing Bernoulli trials).

36 . The group of interest is the cards that are spades, the size of the group of interest is 13, and the sample size is five.

4.6: poisson distribution

37 . A Poisson distribution models the number of events occurring in a fixed interval of time or space, when the events are independent and the average rate of the events is known.

38 . X ~ P (4)

39 . The domain of X = {0, 1, 2, 3, …..) i.e., any integer from 0 upwards.

40 . $\mu =4$
$\sigma =\sqrt{4}=2$

5.1: continuous probability functions

41 . The discrete variables are the number of books purchased, and the number of books sold after the end of the semester. The continuous variables are the amount of money spent for the books, and the amount of money received when they were sold.

42 . Because for a continuous random variable, P ( x = c ) = 0, where c is any single value. Instead, we calculate P ( c < x < d ), i.e., the probability that the value of x is between the values c and d .

43 . Because P ( x = c ) = 0 for any continuous random variable.

44 . P ( x >5) = 1 – 0.35 = 0.65, because the total probability of a continuous probability function is always 1.

45 . This is a uniform probability distribution. You would draw it as a rectangle with the vertical sides at 0 and 20, and the horizontal sides at $\frac{1}{10}$ and 0.

46 . $P\left(0<x<4\right)=\left(4-0\right)\left(\frac{1}{10}\right)=0.4$

5.2: the uniform distribution

47 . $P\left(2<x<5\right)=\left(5-2\right)\left(\frac{1}{10}\right)=0.3$

48 . X ~ U (0, 15)

49 . $f(x)=\frac{1}{b-a}$ for $(a\le x\le b)\text{ so }f(x)=\frac{1}{30}$ for $(0\le x\le 30)$

50 . $\mu =\frac{a+b}{2}=\frac{0+30}{5}=15.0$
$\sigma =\sqrt{\frac{{(b-a)}^2}{12}}=\sqrt{\frac{{\left(30-0\right)}^2}{12}}=8.66$

51 . $P\left(x<10\right)=\left(10\right)\left(\frac{1}{30}\right)=0.33$

5.3: the exponential distribution

52 . X has an exponential distribution with decay parameter m and mean and standard deviation $\frac{1}{m}$ . In this distribution, there will be a relatively large numbers of small values, with values becoming less common as they become larger.

53 . $\mu =\sigma =\frac{1}{m}=\frac{1}{10}=0.1$

54 . f ( x ) = 0.2 e ^{–0.2 x} where x ≥ 0.

6.1: the standard normal distribution

55 . The random variable X has a normal distribution with a mean of 100 and a standard deviation of 15.

56 . X ~ N (0,1)

57 . $z=\frac{x-\mu }{\sigma }$ so $z=\frac{112-109}{4.5}=0.67$

58 . $z=\frac{x-\mu }{\sigma }$ so $z=\frac{100-109}{4.5}=-2.00$

59 . $z=\frac{105-109}{4.5}=\text{−0.89}$
This girl is shorter than average for her age, by 0.89 standard deviations.

60 . 109 + (1.5)(4.5) = 115.75 cm

61 . We expect about 68 percent of the heights of girls of age five years and zero months to be between 104.5 cm and 113.5 cm.

62 . We expect 99.7 percent of the heights in this distribution to be between 95.5 cm and 122.5 cm, because that range represents the values three standard deviations above and below the mean.

6.2: using the normal distribution

63 . Yes, because both np and nq are greater than five.
np = (500)(0.20) = 100 and nq = 500(0.80) = 400