Question #49745

After the quizzes are marked, you’re told that the scores on the exam were distributed normally (μ=50, σ=10). You are also told that students with scores in the top 10% will receive a chocolate bar, and students with scores in the top 25% will receive a stick of gum.
a) What is the minimum score needed to receive a chocolate bar?
b)Would a score of 57 earn a stick of gum?

I don't know the steps, so confused!
1

Expert's answer

2014-12-08T08:51:54-0500

Answer on Question #49745 – Math – Statistics and Probability

After the quizzes are marked, you're told that the scores on the exam were distributed normally (μ=50, σ=10). You are also told that students with scores in the top 10% will receive a chocolate bar, and students with scores in the top 25% will receive a stick of gum.

a) What is the minimum score needed to receive a chocolate bar?

b) Would a score of 57 earn a stick of gum?

Solution

We need to convert the normal distribution to the standard normal distribution:


Z=xμσ.Z = \frac {x - \mu}{\sigma}.


a) P(X>x1)=P(Z>z1)=1P(Z<z1)=0.1P(Z<z1)=0.9.P(X > x_{1}) = P(Z > z_{1}) = 1 - P(Z < z_{1}) = 0.1 \rightarrow P(Z < z_{1}) = 0.9.

Then from normal table, z1=2.33z_{1} = 2.33.

So


z1=x1μσx1=μ+z1σ=50+2.3310=73.3.z _ {1} = \frac {x _ {1} - \mu}{\sigma} \rightarrow x _ {1} = \mu + z _ {1} \sigma = 5 0 + 2. 3 3 \cdot 1 0 = 7 3. 3.

Answer: 73.3.

b) P(Z>z)=P(Z>x1μσ)=P(Z>575010)=P(Z>0.7)=1P(Z<0.7).P(Z > z) = P\left(Z > \frac{x_1 - \mu}{\sigma}\right) = P\left(Z > \frac{57 - 50}{10}\right) = P(Z > 0.7) = 1 - P(Z < 0.7).

Then from normal table,


P(Z<0.7)=0.7580.\mathrm {P} (\mathrm {Z} < 0. 7) = 0. 7 5 8 0.


And


P(Z>z)=10.758=0.242<0.25.P (Z > z) = 1 - 0. 7 5 8 = 0. 2 4 2 < 0. 2 5.


Hence a score of 57 would earn a stick of gum.

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