Answer:
a) 0.06% probability that 338 or more companies outsourced some part of their manufacturing process in the past two or three years.
b) 90.15% probability that 285 or more companies outsourced some part of their manufacturing process in the past two or three years.
c) 0.23% probability that 48% or less of these companies outsourced some part of their manufacturing process in the past two or three years.
Explanation:
For questions a and b, the normal approximation to the binomial is used, while for question c, the central limit theorem is used.
Binomial probability distribution
Probability of exactly x successes on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that
,
.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
and standard deviation
54% of companies surveyed outsourced some part of their manufacturing process in the past two to three years.
This means that
555 of these companies are contacted.
This means that
Mean and standard deviation: Normal approximation to the binomial:
a. What is the probability that 338 or more companies outsourced some part of their manufacturing process in the past two or three years?
Using continuity correction, this is
, which is 1 subtracted by the p-value of Z when X = 337.5. So
has a p-value of 0.9994.
1 - 0.9994 = 0.0006
0.0006*100% = 0.06%
0.06% probability that 338 or more companies outsourced some part of their manufacturing process in the past two or three years.
b. What is the probability that 285 or more companies outsourced some part of their manufacturing process in the past two or three years?
Using continuity correction, this is
, which is 1 subtracted by the p-value of Z when X = 284.5. So
has a p-value of 0.0985.
1 - 0.0985 = 0.9015
0.9015*100% = 90.15%
90.15% probability that 285 or more companies outsourced some part of their manufacturing process in the past two or three years.
c. What is the probability that 48% or less of these companies outsourced some part of their manufacturing process in the past two or three years?
Now we use the sampling distribution of the sample proportions, which have:
The probability is the p-value of Z when X = 0.48. So
By the Central Limit Theorem
has a p-value of 0.0023.
0.0023*100% = 0.23%
0.23% probability that 48% or less of these companies outsourced some part of their manufacturing process in the past two or three years.