Question

suppose the time it takes my daugther, lizzie, to eat an apple is uniformly distributed between 4 and 11 minutes. let x = the time, in minutes, it takes lizzie to eat an apple. a. what is the distribution of x? x - u (4,11) please show the following answers to 4 decimal places. b. what is the probability that it takes lizzie less than 3 minutes to finish the next apple? 0.0 c. what is the probability that it takes lizzie at most 6.3 minutes to finish the next apple? d. what is the probability that it takes lizzie between 4.6 minutes and 9.4 minutes to finish the next apple? e. what is the probability that it takes lizzie fewer than 4.6 minutes or more than 9.4 minutes to finish the next apple?

Explanation:

Step1: Recall the formula for uniform distribution probability

For a uniform distribution \(X\sim U(a,b)\), the probability density function is \(f(x)=\frac{1}{b - a}\) for \(a\leq x\leq b\), and the probability \(P(c\leq X\leq d)=\frac{d - c}{b - a}\) when \(a\leq c\leq d\leq b\). Here \(a = 4\), \(b=11\), so \(f(x)=\frac{1}{11 - 4}=\frac{1}{7}\).

Step2: Solve part c

We want to find \(P(X\leq6.3)\). Since \(a = 4\), using the formula \(P(X\leq x)=\frac{x - a}{b - a}\) (for \(a\leq x\leq b\)), substitute \(x = 6.3\), \(a = 4\), \(b = 11\).
\(P(X\leq6.3)=\frac{6.3-4}{11 - 4}=\frac{2.3}{7}\approx0.3286\)

Step3: Solve part d

We want to find \(P(4.6\leq X\leq9.4)\). Using the formula \(P(c\leq X\leq d)=\frac{d - c}{b - a}\), substitute \(c = 4.6\), \(d = 9.4\), \(a = 4\), \(b = 11\).
\(P(4.6\leq X\leq9.4)=\frac{9.4 - 4.6}{11 - 4}=\frac{4.8}{7}\approx0.6857\)

Step4: Solve part e

We know that \(P(X\lt4.6\text{ or }X\gt9.4)=1 - P(4.6\leq X\leq9.4)\) (by the complement rule). Since \(P(4.6\leq X\leq9.4)\approx0.6857\), then \(P(X\lt4.6\text{ or }X\gt9.4)=1-0.6857 = 0.3143\)

Answer:

c. \(0.3286\)
d. \(0.6857\)
e. \(0.3143\)