Please use the standard normal deviation table to solve the followingRound numbers at the end Combined test scores were normally distributed with mean 1496 and standard deviation 348. Find the combined scores that correspond to these percentiles.a) 10th percentile b) 70th percentile c) 92th percentileClick here to view page 1 of the standard normal distribution table.Click here to view page 2 of the standard normal distribution table.a) The combined scores that correspond to 10th percentile is about(Round to the nearest integer as needed.)b) The combined scores that correspond to 70th percentile is about.(Round to the nearest integer as needed.)c) The combined scores that correspond to 92th percentile is about(Round to the nearest integer as needed.) Entries in the table represent area under the curve between z=0 and a positive value ol z. Because of the symmetry ol thecurve, area under the curve between z 0 and a negative value of z are found in a similar manner.Area - Probability- P(0 s xs 2)0.000.010.020.030.040.050.060.070.080.090.00000.0040.0080012001600199.02390279031903590.1.0398043804780517.0557.0596.063606750714075311410.2.0793083208710910.0948.09871026106411030.3.117912171255.129313311368.1406.1443148015170.41554.15911628166417001736.17721808184418790.519151950.1985201920542088Standard Normal Distribution Table 121232157219022240.622572291232423572389242224542486251725490.725802611264226732704273427642794282328520.828812910293929672995302330513078310631330.931593186321232383264328933153340336533891.0.34133438346134853508353135543577359936211.13643366536863708372937493770.3790381038301.238493869388839073925394439623980399740151.34032404940664082.4099.411541314147.416241771.44192420742224236.4251.42654279429243064319Entries in the table represent area under the curve between z = 0 and a posiive value of z. Because of the symmetry of thecurve, area under the curve between z- 0 and a negative value of z are found in a similar manner.Area - Probability0.000.010.020.030.040.050.060.070.080.09= P(0 s xs 2)1.543324345.4357437043824394442945354406441844411.6dx1.744524463.44744484.449545054515.4525454545544564.457345824591459946084616462546331.8464146494656.46644671467846864693469947061.947134719472647324738474447504756476147672.047724778.478347884793.47984803480848124817Standard Normal Distribution Table 22.148214826.48304834.48384842.48464850485448572.24861.486448684871.4875487848814884488748902.348934896.489849014904.490649094911.491349162.44918492049224925492749294931.4932.4934493625.49384940494149434945494649484949495149522.64953495549564957.49594960.4961496249634964496749762.749654966.4968.4969497049714972497349742.84974.497749754982.4977497849794979498049812.94981498249834984498449854985498649863.049874987.4987.49884988.4989.4989498949904990

Question

Please use the standard normal deviation table to solve the followingRound numbers at the end ￼ Combined test scores were normally distributed with mean 1496 and standard deviation 348. Find the combined scores that correspond to these percentiles.a) 10th percentile b) 70th percentile c) 92th percentileClick here to view page 1 of the standard normal distribution table.Click here to view page 2 of the standard normal distribution table.a) The combined scores that correspond to 10th percentile is about(Round to the nearest integer as needed.)b) The combined scores that correspond to 70th percentile is about.(Round to the nearest integer as needed.)c) The combined scores that correspond to 92th percentile is about(Round to the nearest integer as needed.) Entries in the table represent area under the curve between z=0 and a positive value ol z. Because of the symmetry ol thecurve, area under the curve between z 0 and a negative value of z are found in a similar manner.Area - Probability- P(0 s xs 2)0.000.010.020.030.040.050.060.070.080.090.00000.0040.0080012001600199.02390279031903590.1.0398043804780517.0557.0596.063606750714075311410.2.0793083208710910.0948.09871026106411030.3.117912171255.129313311368.1406.1443148015170.41554.15911628166417001736.17721808184418790.519151950.1985201920542088Standard Normal Distribution Table 121232157219022240.622572291232423572389242224542486251725490.725802611264226732704273427642794282328520.828812910293929672995302330513078310631330.931593186321232383264328933153340336533891.0.34133438346134853508353135543577359936211.13643366536863708372937493770.3790381038301.238493869388839073925394439623980399740151.34032404940664082.4099.411541314147.416241771.44192420742224236.4251.42654279429243064319Entries in the table represent area under the curve between z = 0 and a posiive value of z. Because of the symmetry of thecurve, area under the curve between z- 0 and a negative value of z are found in a similar manner.Area - Probability0.000.010.020.030.040.050.060.070.080.09= P(0 s xs 2)1.543324345.4357437043824394442945354406441844411.6dx1.744524463.44744484.449545054515.4525454545544564.457345824591459946084616462546331.8464146494656.46644671467846864693469947061.947134719472647324738474447504756476147672.047724778.478347884793.47984803480848124817Standard Normal Distribution Table 22.148214826.48304834.48384842.48464850485448572.24861.486448684871.4875487848814884488748902.348934896.489849014904.490649094911.491349162.44918492049224925492749294931.4932.4934493625.49384940494149434945494649484949495149522.64953495549564957.49594960.4961496249634964496749762.749654966.4968.4969497049714972497349742.84974.497749754982.4977497849794979498049812.94981498249834984498449854985498649863.049874987.4987.49884988.4989.4989498949904990

Accepted Answer

Mean =μ=1496, standard deviation=σ=348