Question

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities.

It is known that 79% of all new products introduced in grocery stores fail (are taken off the market) within 2 years. If a grocery store chain introduces 63 new products, find the following probabilities. (Round your answers to four decimal places.)

(a) within 2 years 47 or more fail


(b) within 2 years 58 or fewer fail


(c) within 2 years 15 or more succeed


(d) within 2 years fewer than 10 succeed

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities.

Ocean fishing for billfish is very popular in the Cozumel region of Mexico. In the Cozumel region about 45% of strikes (while trolling) resulted in a catch. Suppose that on a given day a fleet of fishing boats got a total of 21 strikes. Find the following probabilities. (Round your answers to four decimal places.)

(a) 12 or fewer fish were caught


(b) 5 or more fish were caught


(c) between 5 and 12 fish were caught

Answer 1

1)

np=63*0.79=49.77 >10

n(1-p) = 63*(1-0.79) >10

so,  it is appropriate to use the normal approximation to the binomial

a)

Binomial to normal approximations with continuity factor                          
Sample size , n =    63                      
Probability of an event of interest, p =   0.79                      
right tailed                          
X ≥   47                      
                          
Mean = np =    49.77                      
std dev ,σ=√np(1-p)=   3.2329                      
                          
P(X ≥   47   ) = P(Xnormal ≥   46.5   )          
                          
Z=(Xnormal - µ ) / σ = (   46.5   -   49.77   ) /   3.2329   =   -1.011
                          
=P(Z ≥   -1.011   ) =    0.8441              

b)

Sample size , n =    63                      
Probability of an event of interest, p =   0.79                      
left tailed                          
X ≤    58                      
                          
Mean = np =    49.77                      
std dev ,σ=√np(1-p)=   3.2329                      
                          
P(X ≤   58   ) = P(Xnormal ≤   58.5   )          
                          
Z=(Xnormal - µ ) / σ = (   58.5   -   49.77   ) /   3.2329   =   2.700
                          
=P(Z≤   2.700   ) =    0.9965              

c)

Sample size , n =    63                      
Probability of an event of interest, p = 1-0.79= 0.21                      
right tailed                          
X ≥   15                      
                          
Mean = np =    13.23                      
std dev ,σ=√np(1-p)=   3.2329                      
                          
P(X ≥   15   ) = P(Xnormal ≥   14.5   )          
                          
Z=(Xnormal - µ ) / σ = (   14.5   -   13.23   ) /   3.2329   =   0.393
                          
=P(Z ≥   0.393   ) =    0.3472              

d)

Sample size , n =    63                      
Probability of an event of interest, p =   0.21                      
left tailed                          
X <   10                      
                          
Mean = np =    13.23                      
std dev ,σ=√np(1-p)=   3.2329                      
                          
P(X <   10   ) = P(Xnormal <   9.5   )          
                          
Z=(Xnormal - µ ) / σ = (   9.5   -   13.23   ) /   3.2329   =   -1.1538
                          
=P(Z<   -1.1538   ) =    0.1243              
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