Chapter 5 Quiz

Assume that random guesses are made for six multiple choice question on a SAT test, so that there are n=6 trials, each with probability of success (correct) given by p=0.6. Find the indicated probability for the number of correct answers. Find the probability that the number x of correct answers is fewer than 4. P(X<4)=_____

P(X<4)=_0.4557_

Based on a survey, assume that 46% of consumers are comfortable having drones deliver their purchases. Suppose that we want to find the probability that when five consumers are randomly selected, exactly three of them are comfortable with delivery by drones. Identify the values of n, x, p, and q. The value of n is _____. The value of x is _____. The value of p is _____. The value of q is _____.

The value of n is _5_. The value of x is _3_. The value of p is _0.46. The value of q is _0.54.

Five males with an​ X-linked genetic disorder have one child each. The random variable x is the number of children among the five who inherit the​ X-linked genetic disorder. Determine whether a probability distribution is given. If a probability distribution is​ given, find its mean and standard deviation. If a probability distribution is not​ given, identify the requirements that are not satisfied. x P(x) 0 0.029 1 0.147 2 0.324 3 0.324 4 0.147 5 0.029 Does the table show a probability distribution? Select all that apply. a) Yes, the table shows a probability distribution. b) No, the sum of all the probabilities is not equal to 1. c) No, the random variable x's number values are not associated with probabilities. d) No, not every probability is between 0 and 1 inclusive. e) No, the random variable x is categorical instead of numerical. Find the mean of the random variable x. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. a) u=_____ child(ren) b) The table does not show a probability distribution. Find the standard deviation of the random variable x. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. a) o=_____ child(ren) b) The table does not show a probability distribution.

Does the table show a probability distribution? Select all that apply. a) Yes, the table shows a probability distribution. Find the mean of the random variable x. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. a) u=_2.5_ child(ren) 0*0.029+1*0.147+2*0.324+3*0.324+4*0.147+5*0.029=2.5 Find the standard deviation of the random variable x. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. a) o=_1.1_ child(ren) (0-2.5)^2*0.029+(1-2.5)^2*0.147+(2-2.5)^2*0.324+(3-2.5)^2*0.324+(4-2.5)^2*0.147+(5-2.5)^2*0.029=1.1

Refer to the accompanying​ table, which describes results from groups of 8 births from 8 different sets of parents. The random variable x represents the number of girls among 8 children. Find the mean and standard deviation for the number of girls in 8 births. Number of Girls x P(x) 0 0.003 1 0.033 2 0.105 3 0.216 4 0.275 5 0.223 6 0.117 7 0.026 8 0.002 The mean is u=_____ The standard deviation is o=_____

The mean is u=_4.0_ 0*0.003+1*0.033+2*0.105+3*0.216+4*0.275+5*0.223+6*0.117+7*0.026+8*0.002=4.0 The standard deviation is o=_1.4_ (0-4)^2*0.003+(1-4)^2*0.033+(2-4)^2*0.105+(3-4)^2*0.216+(4-4)^2*0.275+(5-4)^2*0.223+(6-4)^2*0.117+(7-4)^2*0.026+(8-4)^2*0.002=1.4

Assume that when adults with smartphones are randomly selected, 63% use them in meetings or classes. If 5 adult smartphone users are randomly selected, find the probability that exactly 3 of them use their smartphones in meetings or classes. The probability is _____.

The probability is _0.3423_. binompdf(5,0.63,3)=0.3423

Assume that when adults with smartphones are randomly​ selected, 57% use them in meetings or classes. If 5 adult smartphone users are randomly​ selected, find the probability that at least 2 of them use their smartphones in meetings or classes. The probability is _____.

The probability is _0.8879_. binompdf(5,0.57,2)= binompdf(5,0.57,3)= binompdf(5,0.57,4)= binompdf(5,0.57,5)= Add together = 0.8879

Assume that different groups of couples use a particular method of gender selection and each couple gives birth to one baby. This method is designed to increase the likelihood that each baby will be a​ girl, but assume that the method has no​ effect, so the probability of a girl is 0.5 Assume that the groups consist of 45couples. Complete parts​ (a) through​ (c) below. a) Find the mean and the standard deviation for the numbers of girls in groups of 45 births. The value of the mean is u=_____ The value of the standard deviation is o=_____ b) Use the range rule of thumb to find the values separating results that are significantly low or significantly high. Values of _____ girls are significantly low. Values of _____ girls are significantly high. c) Is the result of 43 girls a result that is significantly high? What does it suggest about the effective ness of the method? The result _____ significantly high, because 43 girls is __________ _____ girls. A result of 43 girls would suggest that the method __________.

a) Find the mean and the standard deviation for the numbers of girls in groups of 45 births. The value of the mean is u=_22.5_ The value of the standard deviation is o=_3.4_ b) Use the range rule of thumb to find the values separating results that are significantly low or significantly high. Values of _15.7_ girls are significantly low. Values of _29.3_ girls are significantly high. c) Is the result of 43 girls a result that is significantly high? What does it suggest about the effective ness of the method? The result _is_ significantly high, because 43 girls is _greater than_ _29.3_ girls. A result of 43 girls would suggest that the method _is effective_.

The accompanying table describes results from groups of 8 births from 8 different sets of parents. The random variable x represents the number of girls among 8 children. Complete parts​ (a) through​ (d) below. Number of Girls x P(x) 0 0.004 1 0.036 2 0.107 3 0.178 4 0.350 5 0.178 6 0.107 7 0.036 8 0.004 a) Find the probability of getting exactly 6 girls in 8 births. _____ b) Find the probability of getting 6 or more girls in 8 births. _____ c) Which probability is relevant for determining whether 6 is significantly high number in 8 births: the result from part (a) or part (b)? A) The result from part b, since it is the probability of the given or more extreme result. B) The result from part a, since it is the exact probability being asked. C) The result from part b, since it is the complement of the result of part a. D) The result from part a, since it is less than the probability of the given or more extreme result. d) Is 6 a significantly high number of girls in 8 births? Why or why not? Use 0.05 as the threshold for a significant event. A) No, since the appropiate probability is greater than 0.05, it is not a significanly high number. B) Yes, since the appropriate probability is greater than​ 0.05, it is a significantly high number. C) No, since the appropriate probability is less than​ 0.05, it is not a significantly high number. D) ​Yes, since the appropriate probability is less than​ 0.05, it is a significantly high number.

a) Find the probability of getting exactly 6 girls in 8 births. _0.107_ P(x) of line 6 =0.107 b) Find the probability of getting 6 or more girls in 8 births. _0.147_ Add P(x) of line 6, 7, & 8 = 0.147 c) Which probability is relevant for determining whether 6 is significantly high number in 8 births: the result from part (a) or part (b)? A) The result from part b, since it is the probability of the given or more extreme result. d) Is 6 a significantly high number of girls in 8 births? Why or why not? Use 0.05 as the threshold for a significant event. A) No, since the appropiate probability is greater than 0.05, it is not a significanly high number.

There is a 0.9987 probability that a randomly selected 33-year-old male lives through the year. A life insurance company charges $190 for insuring that the male will live through the year. If the male does not survive the year, the policy pays out $110000 as a death benefit. Complete parts (a) through (c) below. a) From the perspective of the 33-year-old male, what are the monetary values corresponding to the two events of surving the year and not surviving? The value corresponding to surviving the year is $_____. The value corresponding to not surviving the year is $_____. b) If the 33-year-old male purchases the policy, what is his expected value? The expected value is $_____. c) Can the insurance company expect to make a profit from many such policies? Why? _____ because the insurance company expects to make an average profit of $_____ on every 33-year-old male it insures for 1 year.

a) From the perspective of the 33-year-old male, what are the monetary values corresponding to the two events of surving the year and not surviving? The value corresponding to surviving the year is $_-190_. Amount paid out. =-190 The value corresponding to not surviving the year is $_109810_. $110000-190=$109810 b) If the 33-year-old male purchases the policy, what is his expected value? The expected value is $_-47.00 c) Can the insurance company expect to make a profit from many such policies? Why? _Yes_ because the insurance company expects to make an average profit of $_47.00_ on every 33-year-old male it insures for 1 year.

When playing roulette at a casino, a gambler is trying to decide whether to be $5 on the number 30 or to bet $5 that the outcome is any one of the three possibilities 00, 0, or 1. The gambler knows that the expected value of the $5 bet for single number is -0.26c. For the $5 bet that the outcome is 00, 0, or 1, there is a probability of 3/38 of making a net profit of $15 and a 35/38 probability of losing $5. a) The expected value is $_____. b) Since the expected value of the bet on the number 30 is __________ than the expected value for the bet that the outcome is 00, 0, or 1, the bet on __________ is better.

a) The expected value is $_-3.42_. (3/38)(15)-(35/38)(5)=-3.42 b) Since the expected value of the bet on the number 30 is _greater_ than the expected value for the bet that the outcome is 00, 0, or 1, the bet on _the single number_ is better.