Question

In a large restaurant an average 60% customers ask for water with their meal. A random sample of 10 customers is selected. Find the probability that,
(a) exactly 6 ask for water with their meal(2 points)
(b) less than 9 ask for water with their meal.(2 points)
(c) at least 3 ask for water with their meal.(2 points)
(d) Find the mean and the standard deviation. (3 points)

Answer 1

Answer:

a)$P(6)=0.25$

b)$p(x$

c)$p(x\geq3)=0.9878$

d)$\sigma=\sqrt{2.4}=1.5492$

Explanation:

From the question we are told that:

Population percentage $p_\%=\60%$

Sample size $n=10$

Let x =customers ask for water

Let y =customers dose not ask for water with their meal  

Generally the equation for y is mathematically given by

$y=1-p_\%\\y=1-0.60\\y=0.40$

Generally the equation for pmf p(x) is mathematically given by

$P(x)=10C_x (0..6)^x(0.4)^{10-x}$

a)

Generally the probability that exactly 6 ask for water is mathematically given by

$P(x)6=10C_6 (0..6)^6(0.4)^{10-6}$

$P(6)=0.25$

b)

Generally the probability that  less than 9 ask for water with meal  is mathematically given by

$p(xg)$

$p(x$

$p(x$

$p(x$

c)

Generally the probability that  at least 3 ask for water with meal  is mathematically given by

$p(x\geq3)=1-p(x$

$p(x\geq3)=1-[p(0)+p(1)+p(2)]$

$p(x\geq3)=1-[0.00001+0.0015+0.0106]$

$p(x\geq3)=1-[0.0122]$

$p(x\geq3)=0.9878$

d)

Generally the mean and standard deviation of sample size is mathematically given by

Mean

$\=x=np=10(0.6)=6$

Standard deviation

$v(x)=npq=10(0.6)(0.4)=2.4$

$\sigma=\sqrt{2.4}=1.5492$