Question

score on last try: 8.67 of 13 pts. see details for more.
at least one scored part is incorrect. jump to first changable incorrect part.
next question get a similar question you can retry this question below
in an effort to cut costs and improve profits, any us companies have been turning to outsourcing. in fact, according to purchasing magazine, 54% of companies surveyed outsourced some part of their manufacturing process in the past two to three years. suppose 555 of these companies are contacted.
a. what is the probability percentage that 338 or more companies outsourced some part of their manufacturing process in the past two or three years? round the percent to two decimal places.
0.07
b. what is the probability percentage that 285 or more companies outsourced some part of their manufacturing process in the past two or three years? round the percent to two decimal places.

c. what is the probability percentage that 483 or less of these companies outsourced some part of their manufacturing process in the past two or three years? round the percent to two decimal places.
0.04

Explanation:

To solve part (b), we can use the normal approximation to the binomial distribution. Here, the number of trials \( n = 555 \), the probability of success \( p = 0.54 \), and we want to find \( P(X \geq 285) \).

Step 1: Calculate the mean and standard deviation of the binomial distribution

The mean \( \mu = np \) and the standard deviation \( \sigma = \sqrt{np(1 - p)} \).

• Mean:

\( \mu = 555 \times 0.54 = 299.7 \)

• Standard deviation:

\( \sigma = \sqrt{555 \times 0.54 \times (1 - 0.54)} = \sqrt{555 \times 0.54 \times 0.46} \approx \sqrt{137.862} \approx 11.74 \)

Step 2: Apply the continuity correction

For \( P(X \geq 285) \) in the binomial distribution, we use the continuity correction and find \( P(X \geq 284.5) \) in the normal distribution.

Step 3: Calculate the z-score

The z-score is given by \( z = \frac{x - \mu}{\sigma} \), where \( x = 284.5 \).

$$ z = \frac{284.5 - 299.7}{11.74} = \frac{-15.2}{11.74} \approx -1.29 $$

Step 4: Find the probability using the z-table

We want \( P(Z \geq -1.29) \). Since the total area under the standard normal curve is 1, we can use the property \( P(Z \geq -1.29) = 1 - P(Z < -1.29) \).

From the z-table, \( P(Z < -1.29) \approx 0.0985 \). Thus:

$$ P(Z \geq -1.29) = 1 - 0.0985 = 0.9015 $$

Step 5: Convert to a percentage

Multiply the probability by 100 to get the percentage:
\( 0.9015 \times 100 = 90.15\% \)

Answer:

The probability percentage that 285 or more companies outsourced some part of their manufacturing process is \(\boldsymbol{90.15\%}\) (rounded to two decimal places).