Answer:
[tex]P(x\:<\:50)=0.0918[/tex]
Step-by-step explanation:
To find the probability that a randomly selected test taker scored below 50, we need to first of all determine the z-score of 50.
The z-score for a normal distribution is given by:
[tex]z=\frac{x-\bar x}{\sigma}[/tex].
From the question, the mean score is [tex]\bar x=70[/tex], the standard deviation is, [tex]\sigma=15[/tex], and the test score is [tex]x=50[/tex].
We substitute these values into the formula to get:
[tex]z=\frac{50-70}{15}[/tex].
[tex]z=\frac{-20}{15}=-1.33[/tex].
We now read the area that corresponds to a z-score of -1.33 from the standard normal distribution table.
From the table, a z-score of -1.33 corresponds to and area of 0.09176.
Therefore the probability that a randomly selected test taker scored below 50 is [tex]P(x\:<\:50)=0.0918[/tex]
