The distribution for scores of an assignment is approximately normal with mean 60 and standard deviation 15. For 1 randomly selected score what is the probability that the score is less than 50? For 10 randomly selected scores what is the probability that at least 5 have scores less than 50?
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normal distribution: The distribution for scores of an assig
- Thread starter waxydock
- Start date
I had a go, but i dont think my answers are right.
thanks for any help.
the first bit
\(\displaystyle \( X \sim N (60, 15^2) \)\)
\(\displaystyle \( Z = \frac{(X - 60)}{15} < \frac{(50 - 60)}{15} \)\)
\(\displaystyle = \(P(z < -2/3) \)\)
\(\displaystyle = \(1 - P (z \leq 0.666..)\)\)
\(\displaystyle = \(1 - 0.7454\)\)
\(\displaystyle = \(0.2546 \)\)
i wasnt sure with the second bit, but i had a go.
\(\displaystyle \( X \sim N (60, \frac{15^2}{10}) \)\)
\(\displaystyle \( X \sim N (60*5, \frac{15^2}{10}*5 )\)\)
\(\displaystyle \( X \sim N (300, 112.5 )\)\)
\(\displaystyle \( Z = \frac{(X - 300)}{112.5} < \frac{(50 - 300)}{112.5} )\)\)
\(\displaystyle = \(P(z < -2.222...) \)\)
\(\displaystyle = \(1 - 0.9861\)\)
\(\displaystyle = \(0.0139 \)\)
thanks for any help.
the first bit
\(\displaystyle \( X \sim N (60, 15^2) \)\)
\(\displaystyle \( Z = \frac{(X - 60)}{15} < \frac{(50 - 60)}{15} \)\)
\(\displaystyle = \(P(z < -2/3) \)\)
\(\displaystyle = \(1 - P (z \leq 0.666..)\)\)
\(\displaystyle = \(1 - 0.7454\)\)
\(\displaystyle = \(0.2546 \)\)
i wasnt sure with the second bit, but i had a go.
\(\displaystyle \( X \sim N (60, \frac{15^2}{10}) \)\)
\(\displaystyle \( X \sim N (60*5, \frac{15^2}{10}*5 )\)\)
\(\displaystyle \( X \sim N (300, 112.5 )\)\)
\(\displaystyle \( Z = \frac{(X - 300)}{112.5} < \frac{(50 - 300)}{112.5} )\)\)
\(\displaystyle = \(P(z < -2.222...) \)\)
\(\displaystyle = \(1 - 0.9861\)\)
\(\displaystyle = \(0.0139 \)\)