To determine the percentage of students expected to receive a score lower than 70, we need to find the probability that a student's score falls below 70 in a normal distribution with a mean of 80 and a standard deviation of 10.

Step 1: Standardize the Score

First, we convert the score of 70 into a standard score (also known as a z-score) using the formula:

$z = \frac{X - \mu}{\sigma}$

where:

• $X = 70$ (the score we are interested in),
• $\mu = 80$ (the mean of the distribution),
• $\sigma = 10$ (the standard deviation of the distribution).

Plugging in the values:

$z = \frac{70 - 80}{10} = \frac{-10}{10} = -1$

So, the z-score for a score of 70 is -1.

Step 2: Find the Probability

Next, we use the z-score to find the corresponding probability. The z-score of -1 corresponds to the area to the left of this z-score on the standard normal distribution curve.

Looking up a z-score of -1 in the standard normal distribution table or using a calculator, we find:

$P(Z < -1) \approx 0.1587$

Step 3: Interpret the Result

The probability $0.1587$ means that approximately 15.87% of the students are expected to receive a score lower than 70.

Summary:

• Percentage of students scoring below 70: Approximately 15.87%.

Would you like more details on this process or any other questions related to normal distribution?

Here are 5 relative questions that expand on this topic:

1. What percentage of students is expected to score higher than 90?
2. How would the percentage change if the standard deviation was 5 instead of 10?
3. What score corresponds to the top 10% of the students in the class?
4. What is the probability that a student scores between 70 and 90?
5. How would you calculate the z-score for a score of 85?

Tip: When dealing with normally distributed data, z-scores help convert raw data into a standardized form, making it easier to calculate probabilities and compare different data points.