Question

1.      Current estimates suggest that 72% of the home-based computers in a country have access to on-line services. Suppose 16 people with home-based computers were randomly and independently sampled. Find the probability that fewer than 9 of those sampled currently have access to on-line services. Show your work. (3)

 

Answer =  

 

2.      Given that X is a normally distributed variable with a mean of 52.2 and a standard deviation of 2.3, find the probability that X is between 47 and 54. (1)

 

Answer=

 

 

3.      A machine fills jars with coffee beans. The weight of the beans that the machine puts in a jar follows a normal distribution with the mean 246 grams and the standard deviation 3.5 grams. The label on a jar says, “weight 240 grams.” Two filled jars are selected at random. Find the probability that at least one of them is under-filled (has less than 240 grams of beans in it). Show your work. (4)

Answer 1

1) We have that

p = 0.72

n = 16

m = 9

Need to find $P(X<9) = 1 – P(X\ge9)$

$P(X\ge9)=P(X=9)+P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)$

This follows binomial distribution.

The binomial probability is calculated by the formula:

$$P(X=m)=C(n,m)\cdot p^m \cdot (1-p)^{n-m}$$

$P(X=9)=C(16,9)\cdot 0.72^9 \cdot (1-0.72)^{16-9}=\frac{16!}{7!9!}\cdot0.72^9\cdot0.28^{7}=0.08$

$P(X=10)=C(16,10)\cdot 0.72^{10} \cdot 0.28^6=0.144$

$P(X=11)=C(16,11)\cdot 0.72^{11} \cdot 0.28^5=0.203$

$P(X=12)=C(16,12)\cdot 0.72^{12} \cdot 0.28^4=0.217$

$P(X=13)=C(16,13)\cdot 0.72^{13} \cdot 0.28^3=0.172$

$P(X=14)=C(16,14)\cdot 0.72^{14} \cdot 0.28^2=0.095$

$P(X=15)=C(16,15)\cdot 0.72^{15} \cdot 0.28^1=0.032$

$P(X=16)=C(16,16)\cdot 0.72^{16} \cdot 0.28^0=0.005$

$P(X<1) = 1-0.08-0.144-0.203-0.217-0.172-0.095-0.032-0.005=0.052$

2) We have that

$\mu = 52.2$

$\sigma=2.3$

$P(47<X<54)=P(\frac{47-\mu}{\sigma}<Z<\frac{54-\mu}{\sigma})=P(\frac{47-52.2}{2.3}<Z<\frac{54-52.2}{2.3})=$

$=P(-2.26<Z<0.78)=P(Z<0.78)-P(Z<-2.26)=$

$=0.7823-0.119=0.6633$

A machine fills jars with coffee beans. The weight of the beans that the machine puts in a jar follows a normal distribution with the mean 246 grams and the standard deviation 3.5 grams. The label on a jar says, “weight 240 grams.” Two filled jars are selected at random. Find the probability that at least one of them is under-filled (has less than 240 grams of beans in it). Show your work.

3) We have that

$\mu = 246$

$\sigma=3.5$

$x=240$

$n = 2$

$m=1$

$P(X<240)=P(Z<\frac{x-\mu}{\sigma})=P(Z<\frac{240-246}{3.5})=P(Z<-1.71)=0.0436$

Thus p = 0.0436

At least one of two jars is under-filled:

$P(X\ge1)=1-P(X=0)=1-C(2,0)\cdot 0.0436^{0} \cdot 0.9564^2=0.0853$

Answer:

1) 0.052

2) 0.6633

3) 0.0853