Respuesta :
Complete 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 81% of all new products introduced in grocery stores fail (are taken off the market) within 2 years. If a grocery store chain introduces 60 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
Answer:
a
[tex]P(X > 47.5 ) = 0.64319[/tex]
b
[tex]P(X < 57.5 ) = 0.9985[/tex]
c
[tex]P(X \le 45 ) = 0.091759[/tex]
d
[tex]P(X > 50.5 ) = 0.26337[/tex]
Step-by-step explanation:
From the question we are told that
The proportion of all new products that fail within 2 years is p = 0.81
The sample size is n = 60
Given that this is follows a binomial distribution as stated in the question we have that the mean is
[tex]\mu = np[/tex]
=> [tex]\mu = 60 * 0.81[/tex]
=> [tex]\mu = 48.6[/tex]
Generally the standard deviation is mathematically represented as
[tex]\sigma = \sqrt{n * p * (1-p)}[/tex]
=> [tex]\sigma = \sqrt{60 * 0.81 * (1-0.81)}[/tex]
=> [tex]\sigma =3[/tex]
Considering question a
Generally the probability that within 2 years 47 or more fail is mathematically evaluated using normal approximation as follows
[tex]P(X > 47 ) = P(\frac{X - \mu}{\sigma } > \frac{47 - 48.6}{3} )[/tex]
Now applying continuity correction we have
[tex]P(X > (47+0.5) ) = P(\frac{X - \mu}{\sigma } > \frac{(47+0.5) - 48.6}{3} )[/tex]
=> [tex]P(X > 47.5 ) = P(\frac{X - \mu}{\sigma } >-0.367 )[/tex]
[tex]\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )[/tex]
=> [tex]P(X > 47.5 ) = P(Z >-0.367 )[/tex]
From the z table
The area under the normal curve to the right corresponding to -0.367 is
[tex]P(Z >-0.367 ) = 0.64319[/tex]
So
[tex]P(X > 47.5 ) = 0.64319[/tex]
Considering question b
Generally the probability that within 2 years 58 or fewer fail is mathematically evaluated using normal approximation as follows
[tex]P(X < 58 ) = P(\frac{X - \mu}{\sigma } < \frac{58 - 48.6}{3} )[/tex]
Now applying continuity correction we have
[tex]P(X < (58-0.5) ) = P(\frac{X - \mu}{\sigma } < \frac{(58-0.5) - 48.6}{3} )[/tex]
=> [tex]P(X < 57.5 ) = P(\frac{X - \mu}{\sigma } <2.967)[/tex]
[tex]\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )[/tex]
=> [tex]P(X > 57.5 ) = P(Z <-2.967 )[/tex]
From the z table
The area under the normal curve to the right corresponding to -0.367 is
[tex]P(Z <2.967 ) =0.9985[/tex]
So
[tex]P(X < 57.5 ) = 0.9985[/tex]
Considering question c
Generally the probability that within 2 years 15 or more succeed is equivalent to the probability that within 2 years at most (60 - 15 = 45 ) new product will fail
Generally the probability that within 2 years at most 45 new product will fail is mathematically evaluated using normal approximation as follows
[tex]P(X \le 45 ) = 1- P(X > 45 )[/tex]
Here
[tex]P(X > 45 ) = P(\frac{X - \mu}{ \sigma} > \frac{45 -48.6}{3} )[/tex]
Now applying continuity correction we have
[tex]P(X > (45+0.5) ) = P(\frac{X - \mu}{\sigma } > \frac{(45+0.5) - 48.6}{3} )[/tex]
=> [tex]P(X > 45.5 ) = P(\frac{X - \mu}{\sigma } >-1.033 )[/tex]
[tex]\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )[/tex]
=> [tex]P(X > 45.5 ) = P(Z >-1.033 )[/tex]
From the z table
The area under the normal curve to the right corresponding to -1.33 is
[tex]P(Z >-1.33 ) = 0.90824[/tex]
So
[tex]P(X > 45.5 ) = 0.90824[/tex]
Hence
[tex]P(X \le 45 ) = 1- 0.90824[/tex]
=> [tex]P(X \le 45 ) = 0.091759[/tex]
Considering question d
Generally the probability that within 2 years fewer than 10 succeed is equivalent to the probability that within 2 years (60 - 10 = 50 ) or more new product will fail
Generally the probability that within 2 years 50 or more fail is mathematically evaluated using normal approximation as follows
[tex]P(X > 50 ) = P(\frac{X - \mu}{\sigma } > \frac{50 - 48.6}{3} )[/tex]
Now applying continuity correction we have
[tex]P(X > (50+0.5) ) = P(\frac{X - \mu}{\sigma } > \frac{(50+0.5) - 48.6}{3} )[/tex]
=> [tex]P(X > 50.5 ) = P(\frac{X - \mu}{\sigma } >0.633 )[/tex]
[tex]\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )[/tex]
=> [tex]P(X > 50.5 ) = P(Z >0.633 )[/tex]
From the z table
The area under the normal curve to the right corresponding to -0.367 is
[tex]P(Z >0.633 ) = 0.26337[/tex]
So
[tex]P(X > 50.5 ) = 0.26337[/tex]
