Certainly! Let's break down the problem step by step, using the normal approximation to the binomial distribution with a continuity correction.
### Step 1: Understand the problem
We are given that 63% (or 0.63) of all college seniors have a job prior to graduation, and we are taking a random sample of 100 seniors. We need to find the approximate probability that more than 60 have a job prior to graduation.
### Step 2: Define the binomial distribution parameters
- Proportion ([tex]\( p \)[/tex]): 0.63
- Sample size ([tex]\( n \)[/tex]): 100
### Step 3: Calculate the mean and standard deviation of the binomial distribution
- Mean ([tex]\( \mu \)[/tex]): The mean of a binomial distribution is calculated as:
[tex]\[
\mu = n \cdot p
\][/tex]
Substituting the values:
[tex]\[
\mu = 100 \cdot 0.63 = 63.0
\][/tex]
- Standard Deviation ([tex]\( \sigma \)[/tex]): The standard deviation of a binomial distribution is calculated as:
[tex]\[
\sigma = \sqrt{n \cdot p \cdot (1 - p)}
\][/tex]
Substituting the values:
[tex]\[
\sigma = \sqrt{100 \cdot 0.63 \cdot (1 - 0.63)} = \sqrt{100 \cdot 0.63 \cdot 0.37} \approx 4.828
\][/tex]
### Step 4: Continuity correction
Since we are working with a discrete distribution but using a continuous approximation, we need to apply a continuity correction. We want to find the probability that more than 60 (i.e., [tex]\( X > 60 \)[/tex]) seniors have a job. For continuity correction, we consider:
[tex]\[
P(X > 60) \approx P(X \geq 60.5)
\][/tex]
### Step 5: Convert to the standard normal distribution
We need to convert our continuity-corrected value to a z-score:
[tex]\[
z = \frac{x - \mu}{\sigma}
\][/tex]
Substituting [tex]\( x = 60.5 \)[/tex], [tex]\( \mu = 63.0 \)[/tex], and [tex]\( \sigma = 4.828 \)[/tex]:
[tex]\[
z = \frac{60.5 - 63.0}{4.828} \approx -0.518
\][/tex]
### Step 6: Find the probability using the cumulative distribution function (CDF) for the standard normal distribution
The z-score of -0.518 corresponds to the cumulative probability [tex]\( P(Z \leq -0.518) \)[/tex]. However, we need the probability [tex]\( P(Z > -0.518) \)[/tex]:
[tex]\[
P(Z > -0.518) = 1 - P(Z \leq -0.518)
\][/tex]
Using standard normal distribution tables or a computational tool, we find:
[tex]\[
P(Z \leq -0.518) \approx 0.302
\][/tex]
So,
[tex]\[
P(Z > -0.518) = 1 - 0.302 = 0.698
\][/tex]
### Conclusion
Therefore, the approximate probability that more than 60 college seniors out of a sample of 100 have a job prior to graduation is:
[tex]\[
\boxed{0.698}
\][/tex]
