In this problem we have a normal distribution with:
• μ = mean = 100,
,• σ = standard deviation = 15.
(a) The z-score for a value x = 124 is:
[tex]z=\frac{x-\mu}{\sigma}=\frac{124-100}{15}=\frac{24}{15}=1.6.[/tex](b) We must compute the probability that x is greater than 124, which is equivalent to the probability of having z greater than 1.6. Using a table of z-score, we find that:
[tex]P(x>124)=P(z>1.6)=0.0548.[/tex]Answer
• (a), z = 1.6
,• (b), P(x > 124) = 0.0548
