Answer :
Answer:
A minimum score of 70 is required for a student to earn a 4 or a 5.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 60
Standard Deviation, σ = 19
We are given that the distribution of raw scores is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Top 30.4% students received either a 4 or a 5. Let x denotes the score that separates top 30.4% students from lowest 69.6% students
We have to find the value of x such that the probability is 0.304
P(X > x)
[tex]P( X > x) = P( z > \displaystyle\frac{x - 60}{19})=0.304[/tex]
[tex]= 1 -P( z \leq \displaystyle\frac{x - 60}{19})=0.304 [/tex]
[tex]=P( z \leq \displaystyle\frac{x - 60}{19})=0.696 [/tex]
Calculation the value from standard normal z table, we have,
[tex]P(z<0.513) = 0.696[/tex]
[tex]\displaystyle\frac{x - 60}{19} = 0.513\\x = 69.747 \approx 70[/tex]
A minimum score of 70 is required for a student to earn a 4 or a 5.