Respuesta :
Answer:
(c) [tex]f_{X}(x)=\left \{ {{\frac{1}{35-25}=\frac{1}{10};\ 25<X<35} \atop {0;\ Otherwise}} \right.[/tex]
(b) The probability that time taken by Ayesha to prepare breakfast exceeds 33 minutes is 0.20.
(c) The probability that cooking or preparation time is within 2 mins of the mean time is 0.40.
Step-by-step explanation:
The random variable X follows a Uniform (25, 35).
(a)
The probability density function of an Uniform distribution is:
[tex]f_{X}(x)=\left \{ {{\frac{1}{B-A};\ A<X<B} \atop {0;\ Otherwise}} \right.[/tex]
Then the probability density function of the random variable X is:
[tex]f_{X}(x)=\left \{ {{\frac{1}{35-25}=\frac{1}{10};\ 25<X<35} \atop {0;\ Otherwise}} \right.[/tex]
(b)
Compute the value of P (X > 33) as follows:
[tex]P(X>33)=\int\limits^{35}_{33} {\frac{1}{10}} \, dx \\\\=\frac{1}{10}\cdot\int\limits^{35}_{33} {1} \, dx \\\\=\frac{1}{10}\times [x]^{35}_{33}\\\\=\frac{35-33}{10}\\\\=\frac{2}{10}\\\\=0.20[/tex]
Thus, the probability that time taken by Ayesha to prepare breakfast exceeds 33 minutes is 0.20.
(c)
Compute the mean of X as follows:
[tex]\mu=\frac{A+B}{2}=\frac{25+35}{2}=30[/tex]
Compute the probability that cooking or preparation time is within 2 mins of the mean time as follows:
[tex]P(30-2<X<30+2)=P(28<X<32)[/tex]
[tex]=\int\limits^{32}_{28} {\frac{1}{10}} \, dx \\\\=\frac{1}{10}\cdot\int\limits^{32}_{28}{1} \, dx \\\\=\frac{1}{10}\times [x]^{32}_{28}\\\\=\frac{32-28}{10}\\\\=\frac{4}{10}\\\\=0.40[/tex]
Thus, the probability that cooking or preparation time is within 2 mins of the mean time is 0.40.
