Question 1182257
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mu = population mean = 1462
sigma = population standard deviation = 319


z = (x - mu)/sigma
z = (1890-1462)/319
z = <font color=red>1.34 approximately</font>


I would consider this to be <u>not</u> unusual. In other words, it seems fairly likely. Any z score such that {{{-2 <= z <= 2}}} would be considered usual; anything outside this interval is considered unusual. Keep in mind that your teacher may use another interval, so I would check with them about that. Though usually, if a z score is further than 2 standard deviations from the mean, then it's considered unusual.
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