Question: LDL LDL - Z = (x - ) / 105 -53.05 -1.40787643 108 -50.05 -1.328260421 110 -48.05 -1.275183081 113 -45.05 -1.195567072 115 -43.05 -1.142489732 118

LDL LDL - Z = (x - ) /
105 -53.05 -1.40787643
108 -50.05 -1.328260421
110 -48.05 -1.275183081
113 -45.05 -1.195567072
115 -43.05 -1.142489732
118 -40.05 -1.062873723
120 -38.05 -1.009796384
120 -38.05 -1.009796384
122 -36.05 -0.9567190442
124 -34.05 -0.9036417047
125 -33.05 -0.877103035
125 -33.05 -0.877103035
126 -32.05 -0.8505643652
126 -32.05 -0.8505643652
126 -32.05 -0.8505643652
128 -30.05 -0.7974870258
130 -28.05 -0.7444096863
133 -25.05 -0.664793677
139 -19.05 -0.5055616586
152 -6.05 -0.1605589519
156 -2.05 -0.05440427297
170 11.95 0.3171371034
172 13.95 0.3702144429
174 15.95 0.4232917824
180 21.95 0.5825238008
182 23.95 0.6356011403
189 30.95 0.8213718285
189 30.95 0.8213718285
190 31.95 0.8479104983
190 31.95 0.8479104983
192 33.95 0.9009878378
194 35.95 0.9540651772
197 38.95 1.033681186
198 39.95 1.060219856
206 47.95 1.272529214
206 47.95 1.272529214
210 51.95 1.378683893
212 53.95 1.431761233
224 65.95 1.75022527
226 67.95 1.803302609
x (LDL) 6322
= (x) 6322 40 (N) 158.05
37.68086384

Is my computation for the Mean, Standard deviation and Z Scores correct?

Given the mean and standard deviation computed above, what percentage will be included if you were to conduct an intervention program targeting women with high LDL (LDL 130)? Show computation or the use of online calculator.

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