The scores on a national achievement test are normally distributed with a mean of 50 and a standard deviation of 10. Out of a group of 200 students, how many would you expect to score more than 70?

μ=50, σ=10;

For score 70, convert to standardized value
z=(x-μ)/σ=(70-50)/10=2
P(x>70)=P(z>2)
Look up standard normal distribution table to get
P(z<=2)=0.9772, so
P(x>70)=P(z>2)=1-0.9772=0.0228
or the average number of students with score of 70 or more
= P(x>70)*200=4.56 = 5 approx.

To determine the number of students we would expect to score more than 70, we need to find the area under the normal distribution curve to the right of 70.

To do this, we can use a standard normal distribution table or a statistical software, but let's use the standard normal distribution table for this explanation.

First, we need to calculate the z-score for the value 70. The z-score measures how many standard deviations an individual score is from the mean. We use the formula:

z = (x - μ) / σ

where z is the z-score, x is the value we are interested in (70), μ is the mean (50), and σ is the standard deviation (10).

Plugging in the values, we get:

z = (70 - 50) / 10
z = 2

Now, looking up the z-score of 2 in the standard normal distribution table, we find that the area to the left of 2 is approximately 0.9772.

Since we want the area to the right of 2 (the students who score more than 70), we subtract 0.9772 from 1:

area to the right = 1 - 0.9772
area to the right = 0.0228

This means that approximately 0.0228, or 2.28% of the scores will be above 70.

Finally, to find the number of students who are expected to score more than 70, we multiply the percentage by the total number of students:

Number of students = percentage * total number of students
Number of students = 0.0228 * 200
Number of students ≈ 4.56

Therefore, we would expect around 4 or 5 students to score more than 70.

To find the number of students you would expect to score more than 70, we can use the properties of the normal distribution.

Step 1: First, we need to standardize the score of 70 using the formula for z-score:

Z = (X - μ) / σ

Where:
Z is the z-score,
X is the individual score,
μ is the mean,
σ is the standard deviation.

In this case:
X = 70,
μ = 50,
σ = 10.

Plugging in the values, we get:

Z = (70 - 50) / 10
Z = 20 / 10
Z = 2

Step 2: Looking up the z-score of 2 in the standard normal distribution table or using a calculator, we find that the proportion of scores above 70 is approximately 0.0228.

Step 3: To find the number of students expected to score more than 70, we multiply the proportion by the total number of students:

Expected number of students = Proportion * Total number of students
Expected number of students = 0.0228 * 200
Expected number of students = 4.56

Therefore, you would expect around 4 or 5 students out of a group of 200 to score more than 70.