IQ test scores are normally distributed with a mean of 97 and a standard deviation of 11.An individual's IQ score is found to be 123.Find the z-score corresponding to this value
Z = (score-mean)/SD
For a particular value, this table gives the percent of scores between the mean and the z-value of a normally distributed random variable. What percent of the total population is found between the mean and the z-score, assume z = 2.79
For a particular value, this table gives the percent of scores between the mean and the z-value of a normally distributed random variable. What percent of the total population is found between the mean and the z-score, assume z = 2.57.
PLEASE HELP!!!
To find the z-score corresponding to a given value in a normal distribution, you can use the formula:
z = (x - μ) / σ
where:
- z is the z-score
- x is the given value
- μ is the mean of the distribution
- σ is the standard deviation of the distribution
In this case, the given value is the individual's IQ score, x = 123. The mean of the IQ test scores distribution is μ = 97, and the standard deviation is σ = 11.
Plugging in the values, we get:
z = (123 - 97) / 11
z = 26 / 11
z = 2.36 (rounded to two decimal places)
Therefore, the z-score corresponding to an IQ score of 123 is approximately 2.36.