An accountant is studying the attendance levels at the concerts of a touring rock band. She finds that the average number of people attending per performance is 1,508, with a standard deviation of 110. Suppose she wants to study only the attendance at the performances in the lowest 27% for number of attendees. If she only looks at 62 performances, what is the largest number of attendees at a performance in her study? Use the z-table below: 0.000.010.020.030.040.050.060.070.080.09-0.7 0.242 0.239 0.236 0.233 0.230 0.227 0.224 0.221 0.218 0.215|-0.6 0.274 0.271 0.268 0.264 0.261 0.258 0.255 0.251 0.248 0.245-0.5 0.309 0.305 0.302 0.298 0.295 0.291 0.288 0.284 0.281 0.278-0.4 0.345 0.341|0.337 0.334 0.330 0.326 0.323 0.319 0.316 0.312-0.3 0.382 0.378 0.374 0.371 0.367 0.363 0.359 0.356 0.352 0.348-0.2 0.421 0.417 0.413 0.409 0.405 | 0.401 0.397 0.394 0.390 |0.386-0.1 0.460 0.456 0.452 0.448 0.444 0.440 0.436 0.433 0.429 0.425 Round the z-score, o and î to two decimal places.Provide your answer below:z -score :0-=x =

Let x¯ be the mean number of attendees at a performance in her study for 62 performances which follows normal distribution. The given mean is 1508, and standard deviation is 110. That is, μ=1508,σ=110. The attendance at the performances in the lowest 27% for number of attendees.The z-score for lowest 27% is obtained by using the given table as follows. Locate 0.27 in the given table which is not in the table the closest value is 0.271. The z value corresponding to value 0.271 is -0.61. Therefore, the z-score is -0.61.The value of σx¯ is, σx¯=σn=11062=1107.874=13.97 Therefore, the value of σx¯ is 13.97.The value of x¯ is, x¯=μ+zσx¯=1508+-0.6113.97=1508-8.52=1499.48 Therefore, the value of x¯ is 1,499.48.