Math  /  Data & Statistics

QuestionThe starting salaries of new graduates in a particular field are normally distributed with a mean of \52,000andastandarddeviationof$4,500.a)Whatpercentageofgraduatesearnbetween52,000 and a standard deviation of \$4,500. a) What percentage of graduates earn between \47,500 47,500 and $56,500\$ 56,500 ? b) What is the cutoff salary for the top 10\% of earners? c) If a company hires 50 graduates, how many would you expect to have starting salaries below $45,000\$ 45,000 ?

Studdy Solution

STEP 1

What is this asking? We're looking at how new graduate salaries spread out, figuring out how many grads make certain amounts of money. Watch out! Don't mix up percentages and probabilities, and make sure to use the z-table correctly!

STEP 2

1. Percentage of graduates between $47,500\$47,500 and $56,500\$56,500.
2. Cutoff salary for the top 10% of earners.
3. Expected number of graduates with salaries below $45,000\$45,000.

STEP 3

Alright, let's **transform** these salaries into **z-scores**!
Remember, the z-score tells us how far a value is from the mean in terms of standard deviations.
The formula is z=xμσ z = \frac{x - \mu}{\sigma} , where xx is our salary, μ\mu is the mean salary, and σ\sigma is the standard deviation.

STEP 4

For $47,500\$47,500, we have z=$47,500$52,000$4,500=$4,500$4,500=1 z = \frac{\$47,500 - \$52,000}{\$4,500} = \frac{-\$4,500}{\$4,500} = -1 .
This means $47,500\$47,500 is **one standard deviation below** the mean.

STEP 5

For $56,500\$56,500, we get z=$56,500$52,000$4,500=$4,500$4,500=1 z = \frac{\$56,500 - \$52,000}{\$4,500} = \frac{\$4,500}{\$4,500} = 1 .
So, $56,500\$56,500 is **one standard deviation above** the mean!

STEP 6

Now, we look up these z-scores (1-1 and 11) in our **z-table** to find the area under the normal curve between these values.
The area between z=1z = -1 and z=1z = 1 represents the percentage of graduates within this salary range.
The z-table tells us the area to the *left* of a given z-score.
The area to the left of z=1z = 1 is approximately 0.84130.8413, and the area to the left of z=1z = -1 is approximately 0.15870.1587.
So, the area *between* them is 0.84130.1587=0.68260.8413 - 0.1587 = 0.6826, or about **68.26%**.

STEP 7

We want to find the salary that corresponds to the **90th percentile** (since the top 10% earn *more* than this cutoff).
We need to find the z-score that has an area of 0.900.90 to its left in the z-table.
This z-score is approximately 1.281.28.

STEP 8

Now, we use the z-score formula, but this time we're solving for xx (the salary): x=μ+zσ x = \mu + z \cdot \sigma .
Plugging in our values, we get x=$52,000+(1.28)$4,500=$52,000+$5,760=$57,760 x = \$52,000 + (1.28) \cdot \$4,500 = \$52,000 + \$5,760 = \$57,760 .
So, the cutoff salary for the top 10% is approximately **\$57,760**.

STEP 9

First, let's find the **z-score** for $45,000\$45,000: z=$45,000$52,000$4,500=$7,000$4,5001.56 z = \frac{\$45,000 - \$52,000}{\$4,500} = \frac{-\$7,000}{\$4,500} \approx -1.56 .

STEP 10

Looking up this z-score in our z-table, we find the area to the left is approximately 0.05940.0594.
This means about **5.94%** of graduates earn less than $45,000\$45,000.

STEP 11

If a company hires 5050 graduates, we'd expect 500.05942.9750 \cdot 0.0594 \approx 2.97 graduates to have starting salaries below $45,000\$45,000.
Since we can't have fractions of people, we round to the nearest whole number, which is **3 graduates**.

STEP 12

a) Approximately **68.26%** of graduates earn between $47,500\$47,500 and $56,500\$56,500. b) The cutoff salary for the top 10% of earners is approximately **\$57,760**. c) We would expect about **3** out of 50 graduates to have starting salaries below \(\$45,000\).

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