STANDARD NORMAL DISTRIBUTION: Table Values Represent AREA to the LEFT of the Z score.Z.00.01.02.03.04.06.07.08-3.9 .00005 .00005 .00004.00004.00004.00004.00004.00003-3.8.00007 .00007 .00007.00006.00006.00006 .00005.00005-3.7.00010.00009.00009.00008 .00008 .00008.00014.00014.00013.00012.00013 .00012.00019 .00018 .00017,00021.00020.00019.00030 .00029.00028 .00027.00026 .00025.00011 .00010 .00010-3.6 .00016 .00015 .00015-3.5 .00023 .00022 .00022-3.4 .00034.00032 .00031-3.3 .00048 .00047 .00045-3.2 .00069 .00066-3.1 .00097-3.0 .00135.00064.00052.00043 .00042 .00040 .00039 .00038 .00036.00062 .00060 .00058 .00056 .00054.00084 .00082.00118 .00114.00094.00090.00087.00079 .00076 .00074.00131 .00126.00122.00107.00104.00111.00164 .00159 .00154.00212.00149.00144.00205.00199.00280.00272.00391 .00379 .00368-2.9 .00187 .00181 .00175.00169-2.8 .00256 .00248 .00240 .00233 ,00226 .00219-2.7 .00347 .00336 .00326 .00317.00307 .00298 .00289-2.6 .00466 .00453 .00440 .00427 .00415 .00402-2.5 .00621 .00604 .00587 .00570 .00554 .00539 .00523 .00508 .00494-2.4 .00820 .00798 .00776 .00755 .00734 .00714.00695.00676 .00657-2.3 .01072 .01044 .01017 .00990 .00964 .00939 .00914 .00889 .00866.01321.01287.01255 .01222 .01191 .01160 .01130-2.1 .01786 .01743 .01700 .01659 .01618 .01578 .01539 .01500 .01463.02068 .02018 .01970 .01923.02619 .02559 .02500 .02442 .02385.03288 .03216.03144.04006 .03920 .03836 .03754.04947.06057 .05938 .05821-2.2 .01390 .01355-2.0 .02275 .02222.02169.02118.01876-1.9.02872.02807.02743.02680-1.8.03593 .03515.03438.03362-1.704457 .04363.0427204182.0409305155.04846 .04746.04648.05705.07078 .06944.08534 .08379.07353 .07215.08851 .08691.10565.12507 .12302 .12100.14686.14457.17106 .16853.19489.10383 .10204 .10027.11900-1.6 .05480 .05370 .05262.05050-1.5 .06681 .06552 .06426 .06301 .06178-1.4 .08076 .07927 .07780 .07636 .07493-1.3 .09680 .09510 .09342 .09176 .09012-1.2 .11507 .11314 .11123 .10935 .10749-1.1 .13567 .13350 .13136 .12924 .12714-1.0 .15866.15625 .15386 .15151 .14917-0.9 .18406 .18141 .17879 .17619 .17361-0.8 .21186 .20897 .20611 .20327-0.7 .24196 23885 .23576 .23270 .2296527425-0.6.27093 .26763 .26435 .26109-0.5 .30854 .30503.14007.16602 .16354.20045.19766.19215 .18943.22663.22363 .2206521770.25785.25463 25143 .24825.30153 .29806.2946029116.2877428434 .28096-0.4.34458.34090.33724.33360.32997.32636.32276.31918.31561-0.3.38209.37828.37448.37070.36693.36317.35942.35569.35197-0.2.40129.39743.39358 .38974.42074 .41683.41294 .40905 ,40517.46017 .45620 .45224 .44828 .44433-0.1.44038.43644.43251 .42858-0.0.50000.49601 .49202.48803 .48405.48006.47210.47608.46812.05.00004.00006.09.00003.00005.00008.00011.00017.00024.00035.00050.00071.00100.00139.00193.00264.00357.00480.0063900842.01101.01426.01831.02330.03074 .03005 .02938.03673.04551.05592.06811.08226.09853.11702.13786.16109.18673.21476.24510.27760.31207.34827.38591.4246546414 The cost of 5 gallons of ice cream has a standard deviation of 7 dollars with a mean of 39 dollars during the summer.What is the probability that the sample mean would be less than 40.2 dollars if a sample of 188 5-gallon pails is randomly selected? Round your answer to fourdecimal places.AnswerHow to enter your answer (opens in new window)TablesKeyKeyboard Sho

population mean cost=39population standard deviation =7 dollarssample of 5 gallon pails=188The z score probability distribution for sample mean is given byTo determine the probability that the sample mean would be les than 40.2 dollars if a sample of 188 5 gallon pails is randomly selected is given by,from the above table value is 0.00939or from excelThe probability that the sample mean would be les than 40.2 dollars if a sample of 188 5 gallon pails is 0.9906The probability that the sample mean would be les than 40.2 dollars if a sample of 188 5 gallon pails is 0.9906