Question

Toms of Maine sells toothpaste in a tube advertised to contain 8 ounces. In their continuous production process, the amount of toothpaste put in a tube is normally distributed with a mean of 8.10 ounces and a standard deviation of 0.05 ounces. If the actual capacity of the tubes used is 8.22 ounces, what proportion of the tubes will be filled beyond capacity?

Answer 1

Process output, X ~ Normal(μ=8.10, σ=0.05)

We have to find the probability P(X > 8.22)

P(X > 8.22) = 1 - P(X < 8.22)

= 1 - P{(X - μ)/σ < (8.22 - μ)/σ}

= 1 - P(Z < (8.22 - 8.10)/0.05) Where Z ~ Normal (0,1)

= 1 - P(Z < 2.4)

= 1 - F(2.4)

Check the standard normal table to find the cumulative distribution of z = 2.4 i.e. F(2.4) = 0.99180

STANDARD NORMAL DISTRIBUTION: Table Values Represent AREA to the LEFT of the Z score Z 00 01 03 .04 05 06 07 .08 09 02 52392 0.0 50000 50399 50798 51197 51595 51994 52790 53188 53586 54776 55172 55567 55962 56356 57535 0.1 53983 54380 .56749 57142 58317 61026 0.2 .57926 58706 59095 .59483 59871 .60257 60642 61409 62930 6443 64803 0.3 61791 62172 62552 63307 63683 64058 65173 0.4 .65542 .65910 66276 66640 67003 67364 .67724 68082 .68439 68793 .70540 .70884 71904 0.5 69146 .69497 69847 .70194 .71226 71566 72240 .73237 .74215 0,6 72575 72907 73565 73891 .74537 .74857 75175 75490 .75804 77337 77637 0.7 .76115 .76424 76730 77035 77935 .78230 .78524 .78814 79389 80511 0.8 .79103 79673 .79955 80234 .80785 81057 81327 81594 83398 0.9 81859 .82121 82381 82639 .82894 83147 83646 83891 4 1.0 4 84375 84614 84849 85083 85 90 1.1 86433 86650 86864 87076 87286 87493 87698 .87900 88100 88298 1.2 88493 88686 88877 89065 .89251 89435 89617 89796 .89973 90147 90824 1.3 .90320 90490 90658 90988 91149 91309 91466 91621 .91774 92647 92922 93056 1.4 91924 92073 92220 92364 92507 .92785 93189 93448 93822 94179 1.5 93319 93574 93699 93943 94062 94295 .94408 95053 95352 1.6 94520 94630 94738 .94845 .94950 95154 .95254 95449 95994 96080 1.7 95543 95637 95728 95818 95907 .96164 96246 .96327 96784 1.8 96407 96485 96562 96638 96712 96856 .96926 96995 97062 1.9 .97128 97193 97257 97320 97381 .97441 97500 .97558 97615 97670 97725 97882 97932 98030 2.0 97778 97831 97982 .98077 98124 98169 2.1 98214 .98257 98300 98341 .98382 98422 98461 .98500 .98537 98574 98645 2.2 .98610 98679 98713 98745 98778 98809 98840 98870 98899 98928 99111 2.3 98956 98983 99010 .99036 99061 99086 99134 99158 2.4 2.5 99180 99245 99266 99286 99305 99202 99224 99324 99343 99361 99379 99430 99396 99413 99446 99461 99477 99520 99492 99506

So,

P(X > 8.22) = 1 - F(2.4) = 1 - 0.99180 = 0.0082