I think I know how to answer the following question IF I assume that the distribution is normal - however, since the question doesn't say that it comes from a normal distribution - should I proceed differently?
Question reads: The mean IQ among members of a club is 118 with a standard deviation of 9.46. What percentage of the members had IQ scores of 115 or higher?
My response,66.3% as follows: Find Z-score of 114 since question asks about 115 or higher (Z=-.42), look up .42 in Z Table (16.28) and add to 50% (66.28%).
IQ is a normal gaussian distribution.
Great job! It looks like you know how to approach this question. However, let me explain it step-by-step, considering the assumption of a normal distribution.
To solve this question, you can use the concept of z-scores. A z-score, or standard score, measures how many standard deviations a particular value is from the mean. Here's how you can proceed:
1. Calculate the z-score: To find the z-score corresponding to an IQ score of 115 or higher, use the formula:
z = (x - μ) / σ
where x is the IQ score (115), μ is the mean of the IQ distribution (118), and σ is the standard deviation (9.46).
Plugging the values in, we get:
z = (115 - 118) / 9.46
2. Look up the z-score: Once you have the z-score, you can use a standard normal distribution table (also known as a Z-table) or a statistical software to find the corresponding percentage.
In your case, with a z-score of -0.3166 (approximately), you need to find the area under the curve to the right of this z-score (since you are looking for IQ scores of 115 or higher). Typically, Z-tables provide the area to the left of the z-score, so you'll need to subtract this value from 1 to get the area to the right.
3. Calculate the percentage: The area to the right of the z-score corresponds to the percentage of members with IQ scores of 115 or higher.
So, subtract the value obtained from the Z-table (or software) from 1 and convert it to a percentage.
Based on these steps, you should get the correct answer, which is approximately 66.28%.
Remember, assuming a normal distribution is a reasonable approximation for many real-world scenarios, even if it is not explicitly mentioned in the question.