Respuesta :
Answer:
(a) Expected value = 10.60
Standard deviation = 1.4434
(b) The probability that the weight of a randomly selected bag is no more than 10.8 pounds is 0.54.
(c) The probability that the weight of a randomly selected bag is at least 10.3 pounds is 0.56.
Step-by-step explanation:
Let X = the quantity of soil put in each bag.
The random variable X follows a Uniform distribution.
The range of the values of X are, [8.1, 13.1].
The probability density function of X is:
[tex]f_{X}(x)=\frac{1}{b-a};\ a<X<b,\ a<b[/tex]
(a)
Compute the expected value of X as follows:
[tex]E(X)=\frac{1}{2}(a+b)=\frac{8.1+13.1}{2}=10.6[/tex]
Thus, the expected value of X is 10.60.
Compute the standard deviation of X as follows:
[tex]SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}=\sqrt{\frac{5^{2}}{12}}=1.4434[/tex]
Thus, the standard deviation of X is 1.4434.
(b)
Compute the probability that the weight of a randomly selected bag is no more than 10.8 pounds as follows:
[tex]P(X\leq 10.8)=\int\limits^{10.8}_{8.1} {\frac{1}{13.1-8.1}}\, dx\\=\frac{1}{5} \int\limits^{10.8}_{8.1} {dx}\,\\=\frac{1}{5} |x|^{10.8}_{8.1} \\=\frac{2.7}{5}\\=0.54[/tex]
Thus, the probability that the weight of a randomly selected bag is no more than 10.8 pounds is 0.54.
(c)
Compute the probability that the weight of a randomly selected bag is at least 10.3 pounds as follows:
[tex]P(X\geq 10.3)=\int\limits^{13.1}_{10.3} {\frac{1}{13.1-8.1}}\, dx\\=\frac{1}{5} \int\limits^{13.1}_{10.3} {dx}\,\\=\frac{1}{5} |x|^{13.1}_{10.3} \\=\frac{2.8}{5}\\=0.56[/tex]
Thus, the probability that the weight of a randomly selected bag is at least 10.3 pounds is 0.56.
