The probability that the mean of this sample is less than 105.4 is 14.46%
The z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
[tex]z=\frac{x-\mu}{\sigma/\sqrt{n} } \\\\x=raw\ score, \sigma=standard\ deviation, \mu=mean,n\ is\ sample\ mean\\\\\\Given\ that\ \sigma=45,\mu=124.5,n\ =36,hence:\\\\For\ x<105.4:\\\\z=\frac{105.4-124.9}{45/\sqrt{36} }=-1.06[/tex]
From the normal distribution table, P(x < 105.4) = P(z < -1.06) = 0.1446 = 14.46%
The probability that the mean of this sample is less than 105.4 is 14.46%
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