Question

for questions 7 and 8, draw the normal distribution curve, then answer the questions.

1. a set of 125 golf scores are normally distributed with a mean of 76 and a standard deviation of 3.

a) what percent of the scores are between 67 and 85?
b) what is the probability that a score is no more than 79?
c) about how many scores fell between one standard deviation of the mean?

1. the talk-time battery life of a group of cell phones is normally distributed with a mean of 5 hours and a standard deviation of 15 minutes.

a) what percent of the phones have a battery life of at least 4 hours and 45 minutes?
b) what percent of the phones have a battery life between 4.5 hours and 5.25 hours?
c) what percent of the phones have a battery life less than 5 hours or greater than 5.5 hours?

1. the number of hours that the employees at the grocery store worked last week is normally distributed with a mean of 24 and a standard deviation of 6. if there are 60 total employees, approximately how many worked at least 30 hours last week?
2. the grade point average (gpa) of the students at lakeview high school is normally distributed with a mean of 3.1 and a standard deviation of 0.3. if there are 1800 students enrolled at the school, approximately how many have a gpa between 2.5 and 3.7?

Explanation:

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Question 7 (Golf Scores, $\mu=76$, $\sigma=3$, $n=125$)

Step1: Calculate z-scores for 67,85

$z_{67}=\frac{67-76}{3}=-3$, $z_{85}=\frac{85-76}{3}=3$

Step2: Use empirical rule for $\pm3\sigma$

Percent = 99.7%

Step3: Calculate z-score for 79

$z_{79}=\frac{79-76}{3}=1$

Step4: Find prob for $z\geq1$

Probability = $1 - 0.8413=0.1587$ (15.87%)

Step5: Use empirical rule for $\pm1\sigma$

Percent = 68%

Step6: Calculate number of scores

$\text{Number}=0.68\times125=85$

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Question 8 (Battery Life, $\mu=5$ hrs=300 mins, $\sigma=15$ mins)

Step1: Convert 4h45m to mins: 285 mins

$z_{285}=\frac{285-300}{15}=-1$

Step2: Find prob for $z\geq-1$

Percent = $1 - 0.1587=0.8413$ (84.13%)

Step3: Convert 4.5h=270m, 5.25h=315m

$z_{270}=\frac{270-300}{15}=-2$, $z_{315}=\frac{315-300}{15}=1$

Step4: Find percent between $z=-2,z=1$

Percent = $0.8413 - 0.0228=0.8185$ (81.85%)

Step5: Convert 5.5h=330m, $z_{330}=\frac{330-300}{15}=2$

Step6: Find percent for $z<0$ or $z>2$

Percent = $0.5 + (1 - 0.9772)=0.5228$ (52.28%)

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Question 9 (Employee Hours, $\mu=24$, $\sigma=6$, $n=60$)

Step1: Calculate z-score for 30

$z_{30}=\frac{30-24}{6}=1$

Step2: Find prob for $z\geq1$

Probability = $1 - 0.8413=0.1587$

Step3: Calculate number of employees

$\text{Number}=0.1587\times60\approx9.52\approx10$

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Question 10 (GPA, $\mu=3.1$, $\sigma=0.3$, $n=1800$)

Step1: Calculate z-scores for 2.5,3.7

$z_{2.5}=\frac{2.5-3.1}{0.3}=-2$, $z_{3.7}=\frac{3.7-3.1}{0.3}=2$

Step2: Use empirical rule for $\pm2\sigma$

Percent = 95.4%

Step3: Calculate number of students

$\text{Number}=0.954\times1800=1717.2\approx1717$

Answer:

Question 7

a) 99.7%
b) 15.87%
c) 85 scores

Question 8

a) 84.13%
b) 81.85%
c) 52.28%

Question 9

Approximately 10 employees

Question 10

Approximately 1717 students