Current engineering graduates earn a mean
starting salary of $62 000 in Canada, with a
standard deviation of $2500. Assuming that
the salaries are normally distributed, what
is the probability that a graduate will find
a job with a starting salary of more than
$65 000?
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$65k is 1.2 s.d. above the mean
use a z-score table to find the % (probability) this represents
Thx
To find the probability that a graduate will find a job with a starting salary of more than $65,000, we need to use the concept of standard normal distribution.
First, we need to calculate the z-score for the value of $65,000 using the formula:
z = (X - μ) / σ
Where:
- X is the value of interest ($65,000 in this case)
- μ is the mean ($62,000 in this case)
- σ is the standard deviation ($2,500 in this case)
Substituting the values:
z = (65,000 - 62,000) / 2,500
z = 3,000 / 2,500
z = 1.2
The z-score of 1.2 indicates that $65,000 is 1.2 standard deviations above the mean.
Next, we need to find the probability associated with this z-score using a standard normal distribution table or a statistical calculator. Looking up the z-score of 1.2 in the table, we find that the corresponding probability is approximately 0.8849.
However, since we want to find the probability of earning MORE than $65,000, we need to account for the area to the right of the z-score.
Since the total area under the curve of a normal distribution is 1, we can subtract the probability we just found from 1 to get the probability of earning more than $65,000.
P(X > $65,000) = 1 - 0.8849
P(X > $65,000) = 0.1151
Therefore, the probability that a graduate will find a job with a starting salary of more than $65,000 is approximately 0.1151, or 11.51%.