The area of the mean for a unformly distribution is gievn below
[tex]\text{Mean}=\frac{a+b}{2}[/tex]
Given that a is the minimum or lowest value and b is the maximum or highest value
From the question given,
a= 0; b=4.1
The mean of this distribution is as shown below
[tex]\begin{gathered} \text{Mean}=\frac{0+4.1}{2} \\ \text{Mean}=\frac{4.1}{2} \\ \text{Mean}=2.05 \end{gathered}[/tex]
Hence the mean of this distribution is 2.05
The standard deviation of the uniformly distribution is given by the formula below:
[tex]s_d=\sqrt[]{\frac{(b-a)^2}{12}}[/tex]
The standard deviation is as shown below:
[tex]\begin{gathered} s_d=\sqrt[]{\frac{(4.1-0)^2}{12}} \\ s_d=\sqrt[]{\frac{4.1^2}{12}} \\ s_d=\sqrt[]{\frac{16.81}{12}} \\ s_d=\sqrt[]{1.400833} \\ s_d=1.183568 \\ s_d=1.1836(\text{correct to 4 decimal place)} \end{gathered}[/tex]
Hence, the standard deviation is 1.1836 to the nearest 4 decimal place
The probability of exactly 2.6 is 0. This is because the probability of finding an exact number of a uniform distribution is 0
c. The probability that the round off error for a jumper's distance is exactly 2.6 is
P(x = 2.6) = 0
The probability that the round off error for the distance that a long jumper has jumped is
between 0.7 and 1.2 mm P(0.7 <3 <1.2) is as shown below:
[tex]\begin{gathered} P(0.7
d. Hence, probability that the round off error for the distance that a long jumper has jumped isbetween 0.7 and 1.2 mm P(0.7 <3 <1.2) is 0.1220 correct to four decimal place
The probability that the jump's round off error is greater than 2.72 is P(x > 2.72). This can be solved as shown below
[tex]\begin{gathered} P(x>2.72)=(4.1-2.72)(\frac{1}{4.1-0}) \\ =1.38\times\frac{1}{4.1} \\ =1.38\times0.2439024 \\ =0.336585\approx0.3366(nearrest\text{ 4 decimal place)} \end{gathered}[/tex]
e. Hence, the probability that the jump's round off error is greater than 2.72 is P(x > 2.72) = 0.3366
P(x > 1.7 x > 0.7) is as shown below:
[tex]\begin{gathered} P(x>1.7\text{ /x>0.7)=}\frac{4.1-1.7}{4.1-0.7} \\ =\frac{2.4}{3.4} \\ =0.705882 \\ \approx0.7059(4\text{ decimal place)} \end{gathered}[/tex]
f. Hence, P(x > 1.7 / x > 0.7) = 0.7059 correct to 4 decimal place
The 43rd percentile is
[tex]\begin{gathered} 43\text{ \% of the the maximum value} \\ =\frac{43}{100}\times4.1 \\ =0.43\times4.1 \\ =1.763 \\ 1.7630(4\text{ decimal place)} \end{gathered}[/tex]
g. Hence, the 43rd percile is 1.763
The minimum for the upper quarter is as shown below
[tex]\begin{gathered} \frac{3}{4}of4.1 \\ =0.75\times4.1 \\ =3.075 \end{gathered}[/tex]
h. Hence, the minimum for the upper quarter is 3.075
