Probability problems, set 2
Shaver Manufacturing, Inc. offers dental insurance to its employees. A recent study by the human resource director shows the annual cost per employee per year followed the normal probability distribution, with a mean of $1,280 and a standard deviation of $420 per year.
What fraction of the employees cost more than $1,500 per year for dental expenses?
The interval for x > $1,500 must be standardized using the formula for z
z = [pic]
Since this z value represents the area of the normal curve from the extreme left to 0.524 to the right of the normal curve midpoint. Hence, the answer to the question is:
Z = 1.00 – 0.524 = 0.476 = 0.48
Looking up that z value in the table for area under the normal curve, look for z = 0.4, move to the right to the column marked “8”, which gives:
P = .3156 or 31.56% of the total Shaver Mfg. employees.
Since the question was posed as a fraction, the answer is “nearly one-third”.
What fraction of the employees cost between $1,500 and 2,000 per year?
First, solve for z for each of the amounts.
We know from the above that $1,500 = z of 0.524
Solving for $2,000, we get:
z = [pic]
Next, solve for the probability interval:
P (1.191 ≥ z ≥ 0.524)
Find the corresponding areas under the normal curve:
P =(0.5+0.1170) – (0.5+0.3015) = 0.617 - .8015 = 0.1845 (ignore the sign or just transpose the order in the original P formula)
P= 0.1845 = 18.45% = “Less than one-sixth”
Sorry, my first answer was off.
Estimate the percent that did not have any dental expense.
If you try solving for both z values, you get zero. So use $1 as the cut-off point
z = [pic]
Proceeding:
1-(0.50 - .0024)*2 = 1-(0.4976)*2 = 1-0.9952 = 0.5%
Answer: one-half of 1 percent.
What was the cost for the 10 percent of employees who incurred the highest dental expenses?
[pic]≥ $1,809.20
7.42 The accounting department at Weston Materials, Inc., a national manufacturer of unattached garages, reports that it takes two construction workers a mean of 32 hours and a standard deviation of 2 hours to erect the Red Barn model. Assume the assembly times follow the normal distribution.
1. Determine the Z value for 29 and 34 hours. What percent of the garages take between 32 hours and 34 hours to erect?
|z = 29 hours |z = 34 hours |
|z = [pic] |z = [pic] |
Proceeding and looking up the corresponding values in the z table
P (-1.5 ≥ z ≥ +1.0) = (0.50-0.0668)+(0.5+0.3413) = 0.8413-.4332 = 0.4081 = 40.81%
2. What percent of the garages take between 29 hours and 34 hours to erect?
Question duplicates #1
3. What percent of the garages take 28.7 hours or less to erect?
z = [pic]
P ≤ 1-(0.50 + 1.65) = 1-51.65 = 1 – 0.7422 = 0.2578 = 25.8%
4. Of the garages, 5 percent take how many hours or more to erect?
z = [pic]
X=35.3 hours
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