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?
try this applet instead of looking up your z-scores in a table or chart, you will love it
You can enter the z-scores , or the data as it stands
http://davidmlane.com/hyperstat/z_table.html
The applet is great! But I wonder, given the information I have, can I assume that the distribution is normal? (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?) If it isn't normal - I can't determine the percentage ... or can I?
To find the percentage of members who had IQ scores of 115 or higher, we can use the concept of standard normal distribution.
1. Standardize the IQ score of 115 using the formula: z = (X - µ) / σ
Where:
X = IQ score (115)
µ = mean IQ (118)
σ = standard deviation (9.46)
Plugging in the values:
z = (115 - 118) / 9.46
Calculate:
z ≈ -0.3183
2. Use a standard normal distribution table or calculator to find the corresponding percentage. The table provides the percentage of values below a given z-score, so we need to find the percentage below -0.3183.
Looking up -0.3183 in the standard normal distribution table, we find that it is approximately 0.3745.
3. Calculate the percentage of members with IQ scores of 115 or higher:
Since we want to find the percentage above 115, we subtract the percentage below -0.3183 from 1 (100%):
Percentage = 1 - 0.3745
Calculate:
Percentage ≈ 0.6255
Therefore, approximately 0.6255 or 62.55% of the members had IQ scores of 115 or higher.