Question 1168117: The scores for a standardized test follow a normal distribution with a mean test score (the x with the line over it)=84 and standard deviationσ=6.
a. Approximately 95% of students score between what two numbers?
(
,
)
b. What percent of students score below 87?
%
c. What percent of students score above 95?
%
d. What percent of students score between 70 and 90?
%
e. What percent of students score between 95 and 100?
%
f. What score must a student get higher than to be in the top 10%?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $X$ be the score on the standardized test. We are given that $X$ follows a normal distribution with a mean $\mu = 84$ and a standard deviation $\sigma = 6$.
**a. Approximately 95% of students score between what two numbers?**
For a normal distribution, approximately 95% of the data falls within 2 standard deviations of the mean.
Lower bound = $\mu - 2\sigma = 84 - 2(6) = 84 - 12 = 72$
Upper bound = $\mu + 2\sigma = 84 + 2(6) = 84 + 12 = 96$
Approximately 95% of students score between **(72, 96)**.
**b. What percent of students score below 87?**
First, we need to find the z-score for a score of 87:
$z = \frac{x - \mu}{\sigma} = \frac{87 - 84}{6} = \frac{3}{6} = 0.5$
Now, we need to find the percentage of students scoring below a z-score of 0.5. Looking up the cumulative probability for $z = 0.5$ in a standard normal distribution table or using a calculator, we find $P(Z < 0.5) \approx 0.6915$.
So, approximately $0.6915 \times 100\% = \boxed{69.15\%}$ of students score below 87.
**c. What percent of students score above 95?**
First, we need to find the z-score for a score of 95:
$z = \frac{x - \mu}{\sigma} = \frac{95 - 84}{6} = \frac{11}{6} \approx 1.83$
Now, we need to find the percentage of students scoring above a z-score of 1.83. This is $P(Z > 1.83) = 1 - P(Z \le 1.83)$. Looking up the cumulative probability for $z = 1.83$ in a standard normal distribution table or using a calculator, we find $P(Z \le 1.83) \approx 0.9664$.
So, $P(Z > 1.83) = 1 - 0.9664 = 0.0336$.
Approximately $0.0336 \times 100\% = \boxed{3.36\%}$ of students score above 95.
**d. What percent of students score between 70 and 90?**
First, we need to find the z-scores for scores of 70 and 90:
For $x = 70$: $z_1 = \frac{70 - 84}{6} = \frac{-14}{6} \approx -2.33$
For $x = 90$: $z_2 = \frac{90 - 84}{6} = \frac{6}{6} = 1$
Now, we need to find the percentage of students scoring between z-scores of -2.33 and 1. This is $P(-2.33 < Z < 1) = P(Z < 1) - P(Z \le -2.33)$. Looking up the cumulative probabilities:
$P(Z < 1) \approx 0.8413$
$P(Z \le -2.33) \approx 0.0099$
So, $P(-2.33 < Z < 1) = 0.8413 - 0.0099 = 0.8314$.
Approximately $0.8314 \times 100\% = \boxed{83.14\%}$ of students score between 70 and 90.
**e. What percent of students score between 95 and 100?**
First, we need the z-score for a score of 95 (which we found in part c, $z_1 \approx 1.83$) and the z-score for a score of 100:
For $x = 100$: $z_2 = \frac{100 - 84}{6} = \frac{16}{6} \approx 2.67$
Now, we need to find the percentage of students scoring between z-scores of 1.83 and 2.67. This is $P(1.83 < Z < 2.67) = P(Z < 2.67) - P(Z \le 1.83)$. Looking up the cumulative probabilities:
$P(Z < 2.67) \approx 0.9962$
$P(Z \le 1.83) \approx 0.9664$
So, $P(1.83 < Z < 2.67) = 0.9962 - 0.9664 = 0.0298$.
Approximately $0.0298 \times 100\% = \boxed{2.98\%}$ of students score between 95 and 100.
**f. What score must a student get higher than to be in the top 10%?**
We need to find the score $x$ such that the area to the right of $x$ under the normal curve is 0.10. This means the area to the left of $x$ is $1 - 0.10 = 0.90$. We need to find the z-score corresponding to a cumulative probability of 0.90.
Looking up the z-value for a cumulative probability of 0.90 in a standard normal distribution table or using a calculator, we find $z \approx 1.28$.
Now, we convert this z-score back to the original score scale:
$z = \frac{x - \mu}{\sigma}$
$1.28 = \frac{x - 84}{6}$
$1.28 \times 6 = x - 84$
$7.68 = x - 84$
$x = 84 + 7.68 = 91.68$
Rounding to the nearest whole number, a student must score higher than $\boxed{92}$ to be in the top 10%.
Final Answer: The final answer is:
a. (72, 96)
b. 69.15%
c. 3.36%
d. 83.14%
e. 2.98%
f. 92
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