x ~ N(400,64)
1) Find the limits within which the central 95% of the distribution lies.
2) find the interquartile range of the distribution.
384.32, 415.68
1) Well, if we want to find the limits within which the central 95% of the distribution lies, we can use the z-score. The z-score tells us how many standard deviations away from the mean a particular value is.
For a normal distribution, approximately 95% of the values lie within 1.96 standard deviations of the mean. So, to find the limits, we can multiply the standard deviation (which is 8 in this case) by 1.96 and add/subtract that from the mean.
So, the limits would be:
Upper limit = 400 + (1.96 * 8)
Lower limit = 400 - (1.96 * 8)
But hey, remember that these are just approximate limits. The actual limits might vary a little.
2) Ah, the interquartile range! It sounds fancy, but it's just the difference between the third quartile (Q3) and the first quartile (Q1) of a distribution.
Now, to find Q1 and Q3, we need to find the values below which 25% and 75% of the data fall, respectively. To calculate these values, we use the formula:
Q1 = mean - (0.6745 * standard deviation)
Q3 = mean + (0.6745 * standard deviation)
So, the interquartile range is:
Q3 - Q1
And there you have it, your shiny interquartile range! Just remember to use it wisely and not to wear it as a headband at parties.
No idea
To answer both questions, we need to look at the properties of the normal distribution.
For a normal distribution, about 95% of the data falls within two standard deviations of the mean.
1) Limits within which the central 95% of the distribution lies:
Given x ~ N(400, 64), the standard deviation is 8 (since it's the square root of 64), and the mean is 400.
To find the limits within which the central 95% of the distribution lies, we need to calculate the z-scores for the upper and lower boundaries.
The z-score formula is:
z = (x - μ) / σ
For the lower boundary:
z = (x - 400) / 8
Since we want the central 95%, we need to find the z-score that leaves 2.5% in the tail, which corresponds to a z-score of -1.96.
So, -1.96 = (x - 400) / 8
Solving for x:
-1.96 * 8 = x - 400
x = -15.68 + 400
x = 384.32
For the upper boundary:
z = (x - 400) / 8
Again, we want the central 95%, so we need a z-score of 1.96 that leaves 2.5% in the tail.
So, 1.96 = (x - 400) / 8
Solving for x:
1.96 * 8 = x - 400
x = 15.68 + 400
x = 415.68
Therefore, the limits within which the central 95% of the distribution lies are 384.32 and 415.68.
2) Interquartile range (IQR) of the distribution:
The interquartile range (IQR) is a measure of the dispersion of data and is calculated as the difference between the 75th percentile (Q3) and the 25th percentile (Q1) of the distribution.
Since we know x ~ N(400, 64), the first step is to find the values of Q1 and Q3.
To find Q1:
Using the z-score formula, we find that the z-score corresponding to the 25th percentile is approximately -0.674.
-0.674 = (x - 400) / 8
Solving for x:
-0.674 * 8 = x - 400
x = -5.392 + 400
x = 394.608
To find Q3:
Similarly, the z-score corresponding to the 75th percentile is approximately 0.674.
0.674 = (x - 400) / 8
Solving for x:
0.674 * 8 = x - 400
x = 5.392 + 400
x = 405.392
Therefore, Q1 = 394.608 and Q3 = 405.392.
Finally, the interquartile range (IQR) is calculated as the difference between Q3 and Q1:
IQR = Q3 - Q1
IQR = 405.392 - 394.608
IQR ≈ 10.784
So, the interquartile range of the distribution is approximately 10.784.