Question

In the following problem, check that it is appropriate to use the normal approdimation to the binomial. Then use the normal distribution to estimate the requested probabilities. it is known that 76% of all new products introduced in grocery stores fall are taken off the market) within 2 years. If a grocery store chain introduces 70 new products, find the following probabilities. (Round your answers to four decimal places.) (a) within 2 47 or more fail ase within 2 years 58 or fewer fail c) within 2 years 15 or more succeed within 2 years fewer than 10 succeed Need Help? Mesek Submik Ansee Save Progress Practice Another Version

Answer 1

The following condition will be used to check if the normal approximation can be done for binomial distribution

np > 5 and n(1-p) > 5

n = number of samples

p = probability or proportion of fail

It is given that probability or proportion, p = 0.76 and n = 70

Therefore,

np = 70*0.76 = 53.2 which is greater than 5

n(1-p) = 70*(1-0.76) = 70*0.24 = 16.8 which is also greater than 5

Therefore, normal approximation is valid.

In Normal approximation, we calculate the Z-score which is given as

$\dpi{100} Z = \frac{X-\mu}{\sigma}$

where,

$\dpi{100} \mu = np = 70*0.76 = 53.2$

$\dpi{100} Standard\;Deviation,\;\sigma=\sqrt{np(1-p)}=\sqrt{70*0.76(1-0.76)}$$\dpi{100} =\sqrt{70*0.76*0.24}=3.5732$

(a) within 2 years 47 or more fail

After applying continuity correction of - 0.5, we will have X to be 47 - 0.5 = 46.5

Z value for 46.5:

$\dpi{100} Z = \frac{46.5-53.2}{3.5732}=\frac{-6.7}{3.5732}=-1.8751$

Probability that within 2 years 47 or more fail is given as

P(X > 46.5) = P(Z > -1.8751) = 1 - P(Z < -1.8751)

P(Z < -1.8751) is calculated from standard normal table or Excel using the function "NORMSDIST". The input will be z value which is -1.8751. Therefore, the value is calculated as 0.0304. (Type in any of the cell in excel =NORMSDIST(-1.8751))

Hence,

P(X > 46.5) = P(Z > -1.8751) = 1 - 0.0304 = 0.9696.

Probability that within 2 years 47 or more will fail is 0.9696

(b) within 2 years 58 or fewer fail

After applying continuity correction of + 0.5, we will have X to be 58 + 0.5 = 58.5

Z value for 58.5:

$\dpi{100} Z = \frac{58.5-53.2}{3.5732}=\frac{5.3}{3.5732}=1.4833$

Probability that within 2 years 58 or fewer fail is given as

P(X < 58.5) = P(Z < 1.4833)

P(Z < 1.4833) is calculated from standard normal table or Excel using the function "NORMSDIST". The input will be z value which is 1.4833. Therefore, the value is calculated as 0.9310 (Type in any of the cell in excel =NORMSDIST(1.4833))

Hence,

P(X < 58.5) = P(Z < 1.4833) = 0.9310.

Probability that within 2 years 58 or fewer will fail is 0.9310.

Success:

n = number of samples

p = probability or proportion of success

It is given that probability or proportion of fail = 0.76, therefore proportion for success, p = 1 - 0.76 = 0.24

Therefore,

$\dpi{100} \mu = np = 70*0.24 = 16.8$

$\dpi{100} Standard\;Deviation,\;\sigma=\sqrt{n(1-p)}=\sqrt{70*0.24(1-0.24)}$$\dpi{100} =\sqrt{70*0.24*0.76}=3.5732$

(c) within 2 years 15 or more succeed

After applying continuity correction of - 0.5, we will have X to be 15 - 0.5 = 14.5

Z value for 14.5:

$\dpi{100} Z = \frac{14.5-16.8}{3.5732}=\frac{-2.3}{3.5732}=-0.6437$

Probability that within 2 years 15 or more succeed is given as

P(X > 14.5) = P(Z > -0.6437) = 1 - P(Z < -0.6437)

P(Z < -0.6437) is calculated from standard normal table or Excel using the function "NORMSDIST". The input will be z value which is -0.6437. Therefore, the value is calculated as 0.2599. (Type in any of the cell in excel =NORMSDIST(-0.6437))

Hence,

P(X > 14.5) = P(Z > -0.6437) = 1 - P(Z < -0.6437) = 1 - 0.2599 = 0.7401

Probability that within 2 years 15 or more will succeed is 0.7401

(d) within 2 years fewer than 10 succeed

After applying continuity correction of - 0.5, we will have X to be 10 - 0.5 = 9.5

Z value for 9.5:

$\dpi{100} Z = \frac{9.5-16.8}{3.5732}=\frac{-7.3}{3.5732}=-2.4030$

Probability that within 2 years fewer than succeed is given as

P(X < 9.5) = P(Z < -2.0430)

P(Z < -2.0430) is calculated from standard normal table or Excel using the function "NORMSDIST". The input will be z value which is -2.0430. Therefore, the value is calculated as 0.0205. (Type in any of the cell in excel =NORMSDIST(-2.0430))

Hence,

P(X < 10) = P(Z < -2.0430) = 0.0205

Probability that within 2 fewer than 10 will succeed is 0.0205

Final answers:

(a) 0.9696

(b) 0.9310

(c) 0.7401

(d) 0.0205