Since in this case we are
only using the variance of the sample and not the variance of the real population,
therefore we use the t statistic. The formula for the confidence interval is:
<span>CI = X ± t * s / sqrt(n) ---> 1</span>
Where,
X = the sample mean = 84
t = the t score which is
obtained in the standard distribution tables at 95% confidence level
s = sample variance = 12.25
n = number of samples = 49
From the table at 95%
confidence interval and degrees of freedom of 48 (DOF = n -1), the value of t
is around:
t = 1.68
Therefore substituting the
given values to equation 1:
CI = 84 ± 1.68 * 12.25 /
sqrt(49)
CI = 84 ± 2.94
CI = 81.06, 86.94
<span>Therefore at 95% confidence
level, the scores is from 81 to 87.</span>
A.) For the Junior Varsity Team, mean would be the appropriate measure of center since the data is <span>symmetric or well-proportioned while we should use standard deviation for getting the measure of spread since it also measures the center and how far the values are from the mean.
b.) For the Varsity Team, the median would be the appropriate measure of the center since the data is skewed left and not evenly distributed so median could be used since it does not account for outliers while we use IQR or interquartile range in measuring the spread of data since IQR does not account for the data that is skewed. </span>
1. find the area of the two triangles, including the shaded section. the formula for this is a=1/2bh
the height and base are both 24 inches. this is because you add the 10 from the side of the square to the 14 that is given
so:
a=1/2(24)(24)
a=(12)(24)
a=288 sq. inches
since there are two triangles, you would multiply the area by 2
a=2(288)
a=576 sq. inches
now, since you only need the unshaded section, you have to take away the shaded section, which is a square. to do this, you must calculate the area of the square and take it away from the area of both triangles.
a=576-(lw)
a=576- (10)(10)
a=576-100
a=476 sq. inches
that is your answer
