what percentage of women have shoe sizes greater than 11.33 if the mean is 8.27 and the deviation is 1.53?
11.33-8.27=3.06/1.53=2
95/2=47.5
50-47.5=2.5
I have no idea what these numbers are supposed to be.
If you are finding the z-score, your mathematical form is really bad
z-score for 11.33 = (11.33-8.27)/1.53 = 2
I you are using tables, a z-score of 2 has value of .9772
that is 97.72% will have a shoe size below 11.33
which means that .0228 or 2.28% have a shoe size above 11.33
Your best bet for these questions is the webpage
http://davidmlane.com/hyperstat/z_table.html
(never seen any shoe size other than multiple of 1/2's . e.g. 9 1/2 )
To find the percentage of women with shoe sizes greater than 11.33, you can use the standard normal distribution. Here's how you can calculate it:
1. Start by finding the z-score for the shoe size 11.33. The z-score represents how many standard deviations a particular value is from the mean.
z-score = (11.33 - mean) / deviation
In this case, the mean is 8.27, and the deviation is 1.53.
z-score = (11.33 - 8.27) / 1.53
z-score = 2.0065 (rounded to four decimal places)
2. Look up the corresponding cumulative probability for the z-score in a standard normal distribution table. The cumulative probability represents the area under the curve to the left of the z-score.
From the table, you would find that the cumulative probability for a z-score of 2.0065 is approximately 0.9788.
3. Subtract the cumulative probability from 1 to find the probability of having a shoe size greater than 11.33.
Probability = 1 - 0.9788
Probability = 0.0212 (rounded to four decimal places)
4. Finally, convert the probability to a percentage by multiplying by 100.
Percentage = 0.0212 x 100
Percentage = 2.12%
Therefore, approximately 2.12% of women have shoe sizes greater than 11.33.