{"id":2567,"date":"2021-12-17T14:25:27","date_gmt":"2021-12-17T19:25:27","guid":{"rendered":"https:\/\/openbooks.macewan.ca\/rcommander\/?post_type=chapter&p=2567"},"modified":"2023-12-25T00:20:24","modified_gmt":"2023-12-25T05:20:24","slug":"4-7-review-questions","status":"publish","type":"chapter","link":"https:\/\/openbooks.macewan.ca\/introstats\/chapter\/4-7-review-questions\/","title":{"raw":"4.7 Review Questions","rendered":"4.7 Review Questions"},"content":{"raw":"
    \r\n \t
  1. Let X<\/em><\/span> be the number of repair calls an appliance repair shop may receive during an hour. The probability distribution of X<\/em><\/span> is given in the following table:\r\n
    \r\n\r\n\r\n\r\n\r\n\r\n
    x<\/em><\/span><\/th>\r\n0<\/td>\r\n1<\/td>\r\n2<\/td>\r\n<\/tr>\r\n<\/thead>\r\n
    P<\/em>(X<\/em>=x<\/em>)<\/span><\/th>\r\n2a<\/em><\/span><\/td>\r\n2a<\/em><\/span><\/td>\r\na<\/em><\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n
      \r\n \t
    1. Find the value of a<\/em><\/span>.<\/li>\r\n \t
    2. Find the mean of X<\/em><\/span>.<\/li>\r\n \t
    3. Find the standard deviation of X<\/em><\/span>.<\/li>\r\n \t
    4. Are the events \u201creceiving no more than one call\u201d and \u201creceiving two calls\u201d mutually exclusive?<\/li>\r\n \t
    5. Are the events \u201creceiving no more than one call\u201d and \u201creceiving two calls\u201d independent? Explain using calculations.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
    6. Roll a balanced die four times,\r\n
        \r\n \t
      1. Find the probability of observing six at least once.<\/li>\r\n \t
      2. Find the probability of observing six exactly once.<\/li>\r\n \t
      3. Find the probability of observing six two to four times inclusively.<\/li>\r\n \t
      4. How many times do we expect to observe a six?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
      5. An insurance company wants to design a homeowner\u2019s policy for mid-priced homes. From data compiled by the company, it is known that the annual claim amount, X<\/em><\/span>, in thousands of dollars, per homeowner is a random variable with the following probability distribution.\r\n
        \r\n\r\n\r\n\r\n\r\n\r\n
        x<\/em><\/span><\/th>\r\n0<\/td>\r\n10<\/td>\r\n50<\/td>\r\n100<\/td>\r\n200<\/td>\r\n<\/tr>\r\n<\/thead>\r\n
        P<\/em>(X<\/em>=x<\/em>)<\/span><\/th>\r\n0.95<\/td>\r\n0.045<\/td>\r\n0.004<\/td>\r\n0.0009<\/td>\r\na<\/em><\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n
          \r\n \t
        1. Determine the value of a<\/em><\/span>.<\/li>\r\n \t
        2. Find the expected annual claim amount per homeowner.<\/li>\r\n \t
        3. Determine the expected annual claim amount for every 1000 homeowners.<\/li>\r\n \t
        4. How much should the insurance company charge for the annual premium to average a net profit of $50 per policy?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
        5. A sales representative for a tire manufacturer claims that the company's steel-belted radials last at least 35,000 miles. A tire dealer decides to check that claim by testing eight of the tires. If 75% or more of the eight tires he tests last at least 35,000 miles, he will purchase tires from the sales representative. If, in fact, 90% of the steel-belted radials produced by the manufacturer last at least 35,000 miles, what is the probability that the tire dealer will purchase tires from the sales representative?<\/li>\r\n \t
        6. From past experience, the owner of a restaurant knows that, on average, 4% of the parties that make reservations never show. How many reservations can the owner accept and still be at least 80% sure that all parties that make a reservation will show?<\/li>\r\n<\/ol>\r\n \r\n\r\n
          Show\/Hide Answer<\/summary>\r\n
            \r\n \t
          1. \u200c\r\n
              \r\n \t
            1. [latex]2a+2a+a=1\\Longrightarrow 5a=1\\Longrightarrow a=0.2.[\/latex]<\/li>\r\n \t
            2. [latex]P(X=0)=2a=0.4, P(X=1)=2a=0.4, P(x=2)=0.2.[\/latex] [latex]\\mu=\\sum x P(X=x)=0\\times 0.4+1\\times 0.4+2\\times 0.2=0.8.[\/latex]<\/li>\r\n \t
            3. [latex]\\sigma=\\sqrt{\\sum x^2 P(X=x)-\\mu^2}=\\sqrt{0^2\\times 0.4+1^2\\times 0.4+2^2\\times 0.2-0.8^2}=0.74833.[\/latex]<\/li>\r\n \t
            4. Let A<\/em><\/span> be the event of receiving no more than one call and B<\/em><\/span> be the event of receiving two calls. The two events are mutually exclusive since they don\u2019t overlap. That is P<\/em>(A<\/em>&B<\/em>)\u2004=\u20040<\/span>.<\/li>\r\n \t
            5. Let A<\/em><\/span> be the event of receiving no more than one call and B<\/em><\/span> be the event of receiving two calls, then\r\n[latex]P(A)=P(X\\le 1)=P(X=0)+P(X=1)=0.4+0.4=0.8, \\quad P(B)=P(X=2)=0.4.[\/latex]<\/span><\/span>The two events are NOT independent, since [latex]P(A\\&B)=0\\ne P(A)\\times P(B)[\/latex]<\/span>.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
            6. \u200c\r\n
                \r\n \t
              1. Let X<\/em><\/span> be the number of six, then X<\/em><\/span> follows a binomial distribution with n<\/em>\u2004=\u20044<\/span> and [latex]p=P(\\mbox{rolling a six})=\\frac{1}{6}[\/latex]<\/span>. We want\r\n[latex]P(X\\ge 1)=1-P(X=0)=1-_4C_0\\left(\\frac{1}{6}\\right)^0\\left(1-\\frac{1}{6}\\right)^{4-0}=1-0.4823=0.5177.[\/latex]<\/span><\/li>\r\n \t
              2. We want\r\n[latex]P(X=1)=_4C_1\\left(\\frac{1}{6}\\right)^1\\left(1-\\frac{1}{6}\\right)^{4-1}=0.3858.[\/latex]<\/span><\/li>\r\n \t
              3. We want\r\n[latex]\\begin{aligned}\r\nP(2\\le X\\le 4)&=P(X=2)+P(X=3)+P(X=4)\\\\\r\n&=_4C_2\\left(\\frac{1}{6}\\right)^2\\left(1-\\frac{1}{6}\\right)^{4-2}+_4C_3\\left(\\frac{1}{6}\\right)^3\\left(1-\\frac{1}{6}\\right)^{4-3}+_4C_4\\left(\\frac{1}{6}\\right)^4\\left(1-\\frac{1}{6}\\right)^{4-4}\\\\\r\n&=0.1157+0.0154+0.0008=0.1319.\\\\\r\n\\text{OR}&=1-P(X=0)-P(X=1)\\\\\r\n&=1-_4C_0\\left(\\frac{1}{6}\\right)^0\\left(1-\\frac{1}{6}\\right)^{4-0}-_4C_1\\left(\\frac{1}{6}\\right)^1\\left(1-\\frac{1}{6}\\right)^{4-1}\\\\\r\n&=1-0.4823-0.3858=0.1319.\\end{aligned}[\/latex]<\/li>\r\n \t
              4. [latex]\\mu=np=4\\times \\frac{1}{6}=0.6667.[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
              5. \u200c\r\n
                  \r\n \t
                1. \r\n
                  \u200c[latex]\\sum P(X=x)=1\\Longrightarrow 0.95+0.045+0.004+0.0009+a=1\\Longrightarrow a=1-0.9999=0.0001.[\/latex]<\/div><\/li>\r\n \t
                2. [latex]\\mu=\\sum x P(X=x)=0\\times 0.95+10\\times 0.045+50\\times 0.004+100\\times 0.0009+200\\times 0.0001=0.76 (\\mbox{\\$1000}).[\/latex]<\/li>\r\n \t
                3. [latex]1000\\times \\mu=1000\\times 0.76=760 (\\mbox{\\$1000}).[\/latex]<\/li>\r\n \t
                4. [latex]\\mu+50=760+50=\\$810.[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                5. Let X<\/em><\/span> be the number of steel-belted radials lasting at least 35000 miles, then X<\/em><\/span> follows a binomial distribution with n<\/em>\u2004=\u20048<\/span> and p<\/em>\u2004=\u20040.9<\/span>. Since 75% of eight is 0.75\u2005\u00d7\u20058\u2004=\u20046<\/span>, we want\r\n[latex]\\begin{aligned}\r\nP(X\\ge 6)&=P(X=6)+P(X=7)+P(X=8)\\\\\r\n&={}_8C_6(0.9)^6(1-0.9)^2+_8C_7(0.9)^7(1-0.9)^7+_8C_8(0.9)^8(1-0.9)^0\\\\\r\n&=0.1488+0.3826+0.4305=0.9616.\\end{aligned}[\/latex]<\/li>\r\n \t
                6. Since 4% of the parties never show, 96% will show. We want\r\n[latex]0.96^n\\ge 0.8\\Longrightarrow n\\le\\frac{\\ln 0.8}{\\ln 0.96}=5.47\\Longrightarrow n=5.[\/latex]<\/span><\/li>\r\n<\/ol>\r\n<\/details>","rendered":"
                    \n
                  1. Let X<\/em><\/span> be the number of repair calls an appliance repair shop may receive during an hour. The probability distribution of X<\/em><\/span> is given in the following table:\n
                    \n\n\n\n\n\n
                    x<\/em><\/span><\/th>\n0<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n<\/thead>\n
                    P<\/em>(X<\/em>=x<\/em>)<\/span><\/th>\n2a<\/em><\/span><\/td>\n2a<\/em><\/span><\/td>\na<\/em><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n
                      \n
                    1. Find the value of a<\/em><\/span>.<\/li>\n
                    2. Find the mean of X<\/em><\/span>.<\/li>\n
                    3. Find the standard deviation of X<\/em><\/span>.<\/li>\n
                    4. Are the events \u201creceiving no more than one call\u201d and \u201creceiving two calls\u201d mutually exclusive?<\/li>\n
                    5. Are the events \u201creceiving no more than one call\u201d and \u201creceiving two calls\u201d independent? Explain using calculations.<\/li>\n<\/ol>\n<\/li>\n
                    6. Roll a balanced die four times,\n
                        \n
                      1. Find the probability of observing six at least once.<\/li>\n
                      2. Find the probability of observing six exactly once.<\/li>\n
                      3. Find the probability of observing six two to four times inclusively.<\/li>\n
                      4. How many times do we expect to observe a six?<\/li>\n<\/ol>\n<\/li>\n
                      5. An insurance company wants to design a homeowner\u2019s policy for mid-priced homes. From data compiled by the company, it is known that the annual claim amount, X<\/em><\/span>, in thousands of dollars, per homeowner is a random variable with the following probability distribution.\n
                        \n\n\n\n\n\n
                        x<\/em><\/span><\/th>\n0<\/td>\n10<\/td>\n50<\/td>\n100<\/td>\n200<\/td>\n<\/tr>\n<\/thead>\n
                        P<\/em>(X<\/em>=x<\/em>)<\/span><\/th>\n0.95<\/td>\n0.045<\/td>\n0.004<\/td>\n0.0009<\/td>\na<\/em><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n
                          \n
                        1. Determine the value of a<\/em><\/span>.<\/li>\n
                        2. Find the expected annual claim amount per homeowner.<\/li>\n
                        3. Determine the expected annual claim amount for every 1000 homeowners.<\/li>\n
                        4. How much should the insurance company charge for the annual premium to average a net profit of $50 per policy?<\/li>\n<\/ol>\n<\/li>\n
                        5. A sales representative for a tire manufacturer claims that the company’s steel-belted radials last at least 35,000 miles. A tire dealer decides to check that claim by testing eight of the tires. If 75% or more of the eight tires he tests last at least 35,000 miles, he will purchase tires from the sales representative. If, in fact, 90% of the steel-belted radials produced by the manufacturer last at least 35,000 miles, what is the probability that the tire dealer will purchase tires from the sales representative?<\/li>\n
                        6. From past experience, the owner of a restaurant knows that, on average, 4% of the parties that make reservations never show. How many reservations can the owner accept and still be at least 80% sure that all parties that make a reservation will show?<\/li>\n<\/ol>\n

                           <\/p>\n

                          \nShow\/Hide Answer<\/summary>\n
                            \n
                          1. \u200c\n
                              \n
                            1. [latex]2a+2a+a=1\\Longrightarrow 5a=1\\Longrightarrow a=0.2.[\/latex]<\/li>\n
                            2. [latex]P(X=0)=2a=0.4, P(X=1)=2a=0.4, P(x=2)=0.2.[\/latex] [latex]\\mu=\\sum x P(X=x)=0\\times 0.4+1\\times 0.4+2\\times 0.2=0.8.[\/latex]<\/li>\n
                            3. [latex]\\sigma=\\sqrt{\\sum x^2 P(X=x)-\\mu^2}=\\sqrt{0^2\\times 0.4+1^2\\times 0.4+2^2\\times 0.2-0.8^2}=0.74833.[\/latex]<\/li>\n
                            4. Let A<\/em><\/span> be the event of receiving no more than one call and B<\/em><\/span> be the event of receiving two calls. The two events are mutually exclusive since they don\u2019t overlap. That is P<\/em>(A<\/em>&B<\/em>)\u2004=\u20040<\/span>.<\/li>\n
                            5. Let A<\/em><\/span> be the event of receiving no more than one call and B<\/em><\/span> be the event of receiving two calls, then
                              \n[latex]P(A)=P(X\\le 1)=P(X=0)+P(X=1)=0.4+0.4=0.8, \\quad P(B)=P(X=2)=0.4.[\/latex]<\/span><\/span>The two events are NOT independent, since [latex]P(A\\&B)=0\\ne P(A)\\times P(B)[\/latex]<\/span>.<\/li>\n<\/ol>\n<\/li>\n
                            6. \u200c\n
                                \n
                              1. Let X<\/em><\/span> be the number of six, then X<\/em><\/span> follows a binomial distribution with n<\/em>\u2004=\u20044<\/span> and [latex]p=P(\\mbox{rolling a six})=\\frac{1}{6}[\/latex]<\/span>. We want
                                \n[latex]P(X\\ge 1)=1-P(X=0)=1-_4C_0\\left(\\frac{1}{6}\\right)^0\\left(1-\\frac{1}{6}\\right)^{4-0}=1-0.4823=0.5177.[\/latex]<\/span><\/li>\n
                              2. We want
                                \n[latex]P(X=1)=_4C_1\\left(\\frac{1}{6}\\right)^1\\left(1-\\frac{1}{6}\\right)^{4-1}=0.3858.[\/latex]<\/span><\/li>\n
                              3. We want
                                \n[latex]\\begin{aligned} P(2\\le X\\le 4)&=P(X=2)+P(X=3)+P(X=4)\\\\ &=_4C_2\\left(\\frac{1}{6}\\right)^2\\left(1-\\frac{1}{6}\\right)^{4-2}+_4C_3\\left(\\frac{1}{6}\\right)^3\\left(1-\\frac{1}{6}\\right)^{4-3}+_4C_4\\left(\\frac{1}{6}\\right)^4\\left(1-\\frac{1}{6}\\right)^{4-4}\\\\ &=0.1157+0.0154+0.0008=0.1319.\\\\ \\text{OR}&=1-P(X=0)-P(X=1)\\\\ &=1-_4C_0\\left(\\frac{1}{6}\\right)^0\\left(1-\\frac{1}{6}\\right)^{4-0}-_4C_1\\left(\\frac{1}{6}\\right)^1\\left(1-\\frac{1}{6}\\right)^{4-1}\\\\ &=1-0.4823-0.3858=0.1319.\\end{aligned}[\/latex]<\/li>\n
                              4. [latex]\\mu=np=4\\times \\frac{1}{6}=0.6667.[\/latex]<\/li>\n<\/ol>\n<\/li>\n
                              5. \u200c\n
                                  \n
                                1. \n
                                  \u200c[latex]\\sum P(X=x)=1\\Longrightarrow 0.95+0.045+0.004+0.0009+a=1\\Longrightarrow a=1-0.9999=0.0001.[\/latex]<\/div>\n<\/li>\n
                                2. [latex]\\mu=\\sum x P(X=x)=0\\times 0.95+10\\times 0.045+50\\times 0.004+100\\times 0.0009+200\\times 0.0001=0.76 (\\mbox{\\$1000}).[\/latex]<\/li>\n
                                3. [latex]1000\\times \\mu=1000\\times 0.76=760 (\\mbox{\\$1000}).[\/latex]<\/li>\n
                                4. [latex]\\mu+50=760+50=\\$810.[\/latex]<\/li>\n<\/ol>\n<\/li>\n
                                5. Let X<\/em><\/span> be the number of steel-belted radials lasting at least 35000 miles, then X<\/em><\/span> follows a binomial distribution with n<\/em>\u2004=\u20048<\/span> and p<\/em>\u2004=\u20040.9<\/span>. Since 75% of eight is 0.75\u2005\u00d7\u20058\u2004=\u20046<\/span>, we want
                                  \n[latex]\\begin{aligned} P(X\\ge 6)&=P(X=6)+P(X=7)+P(X=8)\\\\ &={}_8C_6(0.9)^6(1-0.9)^2+_8C_7(0.9)^7(1-0.9)^7+_8C_8(0.9)^8(1-0.9)^0\\\\ &=0.1488+0.3826+0.4305=0.9616.\\end{aligned}[\/latex]<\/li>\n
                                6. Since 4% of the parties never show, 96% will show. We want
                                  \n[latex]0.96^n\\ge 0.8\\Longrightarrow n\\le\\frac{\\ln 0.8}{\\ln 0.96}=5.47\\Longrightarrow n=5.[\/latex]<\/span><\/li>\n<\/ol>\n<\/details>\n","protected":false},"author":19,"menu_order":7,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2567","chapter","type-chapter","status-publish","hentry"],"part":535,"_links":{"self":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/2567","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/users\/19"}],"version-history":[{"count":20,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/2567\/revisions"}],"predecessor-version":[{"id":4752,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/2567\/revisions\/4752"}],"part":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/parts\/535"}],"metadata":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/2567\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/media?parent=2567"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapter-type?post=2567"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/contributor?post=2567"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/license?post=2567"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}