In a standardized IQ test, scores are normally distributed, with ...

6. Normal Distribution and Continuous Random Variables

Standard Normal Distribution

Statistics

6. Normal Distribution and Continuous Random Variables

Standard Normal Distribution

2:21 minutes

Problem 5.Q.5

Textbook Question

In a standardized IQ test, scores are normally distributed, with a mean score of 100 and a standardized deviation of 15. Use this information in Exercises 3–10. (Adapted from 123test)

What percent of the IQ scores are greater than 112?

Verified step by step guidance

Identify the key parameters of the normal distribution: the mean (μ = 100) and the standard deviation (σ = 15). The problem asks for the percentage of IQ scores greater than 112.

Standardize the raw score of 112 to a z-score using the formula:

z = \frac{x - μ}{σ}

, where x is the raw score, μ is the mean, and σ is the standard deviation.

Substitute the values into the z-score formula:

z = \frac{112 - 100}{15}

. Simplify the numerator and divide by the standard deviation to calculate the z-score.

Use a z-table or a standard normal distribution calculator to find the cumulative probability corresponding to the calculated z-score. This gives the probability of a score being less than 112.

To find the percentage of scores greater than 112, subtract the cumulative probability from 1 (i.e.,

1 - P

, where P is the cumulative probability). Multiply the result by 100 to express it as a percentage.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Normal Distribution

Normal distribution is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In the context of IQ scores, this means that most individuals score around the average (100), with fewer individuals scoring significantly higher or lower.

Z-Score

A Z-score indicates how many standard deviations an element is from the mean. It is calculated by subtracting the mean from the score and then dividing by the standard deviation. For an IQ score of 112, the Z-score helps determine its position relative to the mean and is essential for finding the percentage of scores above this value.

Percentile Rank

Percentile rank is a measure used to indicate the value below which a given percentage of observations fall. In this case, calculating the percentile rank for an IQ score of 112 allows us to determine the percentage of individuals who scored lower than this score, which can then be used to find the percentage of scores that are greater.

Watch next

Master Finding Standard Normal Probabilities using z-Table with a bite sized video explanation from Patrick

Start learning

����app

In a standardized IQ test, scores are normally distributed, with a mean score of 100 and a standardized deviation of 15. Use this information in Exercises 3–10. (Adapted from 123test)What percent of the IQ scores are greater than 112?