Question

There is a 14 percent chance that a Noodles & Company customer will order bread with the meal. Use Excel to find the probability that in a sample of 10 customers (a) More than five will order bread. (Round your answer to 5 decimal places.) P(X > 5) (b) No more than two will. (Round your answer to 4 decimal places.) P(X ≤ 2) (c) None of the 10 will order bread. (Round your answer to 4 decimal places.) P(X = 0)

Answer 1

Answer:

(a) More than five will order bread. P(X > 5)=0.0094

(b) No more than two will. P(X≤2)=0.8455

(c) None of the 10 will order bread. P(X=0)=0.2213

Step-by-step explanation:

In this problem we should use a binomial distribution with p=0.14 and n=10.

a) P(X>5)

$P(X>5)=1-(P(0)+P(1)+P(2)+P(3)+P(4)+P(5))\\\\\\P(0)=\binom{10}{0}p^{0}(1-p)^{10}=1*1*0.2213=0.2213\\\\P(1)=\binom{10}{1}p^{1}(1-p)^{9}=10*0.14*0.25733=0.36026\\\\P(2)=\binom{10}{2}p^{2}(1-p)^{8}=45*0.0196*0.29922=0.26391\\\\P(3)=\binom{10}{3}p^{3}(1-p)^{7}=120*0.00274*0.34793=0.11457\\\\P(4)=\binom{10}{4}p^{4}(1-p)^{6}=210*0.00038*0.40457=0.03264\\\\P(5)=\binom{10}{5}p^{5}(1-p)^{5}=252*0.00005*0.47043=0.00638\\\\\\P(X>5)=1-(0.2213+0.36026+0.26391+0.11457+0.03264+0.00638)\\\\P(X>5)=1-0.99906=0.00094$

b) P(X≤2)

$P(X\leq2)=P(0)+P(1)+P(2)\\\\P(X\leq2)=0.22130+0.36026+0.26391=0.84547$

c) P(X=0)

$P(X=0)=P(0)=0.2213$

Answer 2

Answer:

(a) P(X > 5) = 0.00095

(b) P(X $\leq$ 2) = 0.8455

(c) P(X = 0) = 0.2213

Step-by-step explanation:

We are given that there is a 14 percent chance that a Noodles & Company customer will order bread with the meal.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 10 customers

            r = number of success

           p = probability of success which in our question is % that a Noodles  

                  & Company customer will order bread with the meal, i.e., 14%

LET X = Number of customer that will order bread

Also, it is given that a sample of 26 customers is taken,

So, it means X ~ $Binom(n=10, p=0.14)$

(a) Probability that more than five customer will order bread = P(X > 5)

    P(X > 5) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

= $\binom{10}{6}0.14^{6} (1-0.14)^{10-6} + \binom{10}{7}0.14^{7} (1-0.14)^{10-7} +\binom{10}{8}0.14^{8} (1-0.14)^{10-8} +\binom{10}{9}0.14^{9} (1-0.14)^{10-9} +\binom{10}{10}0.14^{10} (1-0.14)^{10-10}$

= $210 \times 0.14^{6} \times 0.86^{4} +120 \times 0.14^{7} \times 0.86^{3} +45 \times 0.14^{8} \times 0.86^{2} +10 \times 0.14^{9} \times 0.86^{1} +1 \times 0.14^{10} \times 0.86^{0}$

= $9.505 \times 10^{-4}$ = 0.00095

(b) Probability that no more than two customer will order bread = P(X $\leq$ 2)

    P(X $\leq$ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= $\binom{10}{0}0.14^{0} (1-0.14)^{10-0} + \binom{10}{1}0.14^{1} (1-0.14)^{10-1}+\binom{10}{2}0.14^{2} (1-0.14)^{10-2}$

= $1 \times 1 \times 0.86^{10} +10 \times 0.14 \times 0.86^{9} +45 \times 0.14^{2} \times 0.86^{8}$

= 0.8455

(c) Probability that None of the 10 will order bread = P(X = 0)                  

    P(X = 0) = $\binom{10}{0}0.14^{0} (1-0.14)^{10-0}$

                  = $1 \times 1 \times 0.86^{10}$ = 0.2213