Three Rivers publishes a catalog each year, last year it had 198 pages back and front. We
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Answer:
And we can find this probability with this difference:
And we can find the real value with the following excel code using the binomial distribution:
"=BINOM.DIST(110,400,0.2,TRUE)-BINOM.DIST(89,400,0.2,TRUE)"
And we got 0.118 a very close value from the value obtained using the normal approximation
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Let X the random variable of interest, on this case we now that:
The probability mass function for the Binomial distribution is given as:
Where (nCx) means combinatory and it's given by this formula:
We need to check the conditions in order to use the normal approximation.
So we see that we satisfy the conditions and then we can apply the approximation.
If we appply the approximation the new mean and standard deviation are:
So then we can approximate the random variable as
And we want this probability:
We can use the z score formula given by:
And replacing we got:
And we can find this probability with this difference:
And we can find the real value with the following excel code using the binomial distribution:
"=BINOM.DIST(110,400,0.2,TRUE)-BINOM.DIST(89,400,0.2,TRUE)"
And we got 0.118 a very close value from the value obtained using the normal approximation
Answer:
(a) The probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.
(b) The probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.
(c) The probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.
Step-by-step explanation:
The complete question is:
A really bad carton of 18 eggs contains 8 spoiled eggs. An unsuspecting chef picks 5 eggs at random for his “Mega-Omelet Surprise.” Find the probability that the number of unspoiled eggs among the 5 selected is
(a) exactly 5
(b) 2 or fewer
(c) more than 1.
Let <em>X</em> = number of unspoiled eggs in the bad carton of eggs.
Of the 18 eggs in the bad carton of eggs, 8 were spoiled eggs.
The probability of selecting an unspoiled egg is:
A randomly selected egg is unspoiled or not is independent of the others.
It is provided that a chef picks 5 eggs at random.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.556.
The success is defined as the selection of an unspoiled egg.
The probability mass function of <em>X</em> is given by:
(a)
Compute the probability that of the 5 eggs selected exactly 5 are unspoiled as follows:
Thus, the probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.
(b)
Compute the probability that of the 5 eggs selected 2 or less are unspoiled as follows:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
Thus, the probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.
(c)
Compute the probability that of the 5 eggs selected more than 1 are unspoiled as follows:
P (X > 1) = 1 - P (X ≤ 1)
= 1 - P (X = 0) - P (X = 1)
Thus, the probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.