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.

image of normal distribution curve with x-axis labels 67, 70, 73, 76, 79, 82, 85 and percentages 0.1, 2.2, 13.6, 34.1, 34.1, 13.6, 2.2, 0.1
a) what percent of the scores are between 67 and 85?
.1 + 2.2 + 13.6 + 34.1 + 34.1 + 13.6 + 2.2 + .1 = 100
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.

image of normal distribution curve with x-axis labels 4.25, 4.50, 4.75, 5, 5.25, 5.50, 5.75 and percentages 0.1, 2.2, 13.6, 34.1, 34.1, 13.6, 2.2, 0.1
a) what percent of the phones have a battery life of at least 4 hours and 45 minutes?
4.75
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?

© gina wilson (all things algebra)

Explanation:

Question 7a

Step1: Identify the z - scores

The mean $\mu = 76$, standard deviation $\sigma=3$. For $x = 67$, $z=\frac{67 - 76}{3}=\frac{- 9}{3}=- 3$. For $x = 85$, $z=\frac{85 - 76}{3}=\frac{9}{3}=3$.

Step2: Use the empirical rule for normal distribution

The empirical rule states that for a normal distribution, the percentage of data between $z=- 3$ and $z = 3$ is approximately $0.1+2.2 + 13.6+34.1+34.1+13.6+2.2+0.1$. But we can also recall that the total percentage within $z=-3$ to $z = 3$ is about $99.7\%$? Wait, no, looking at the graph, the intervals are: from 67 (z=-3) to 70 (z=-2): 0.1 + 2.2, 70 to 73 (z=-1):13.6, 73 to 76 (z = 0):34.1, 76 to 79 (z = 1):34.1, 79 to 82 (z = 2):13.6, 82 to 85 (z = 3):2.2+0.1. Wait, no, let's sum the percentages between 67 and 85. The regions are: 0.1 (left of 67), 2.2 (67 - 70), 13.6 (70 - 73), 34.1 (73 - 76), 34.1 (76 - 79), 13.6 (79 - 82), 2.2 (82 - 85), 0.1 (right of 85). So between 67 and 85, we sum 2.2+13.6 + 34.1+34.1+13.6+2.2. Let's calculate: 2.2+2.2 = 4.4; 13.6+13.6 = 27.2; 34.1+34.1 = 68.2; then 4.4+27.2=31.6; 31.6 + 68.2=99.8? Wait, maybe the graph has the percentages: from the left, 0.1, 2.2, 13.6, 34.1, 34.1, 13.6, 2.2, 0.1. So between 67 (which is at z=-3, the left end of the 0.1 + 2.2 region) and 85 (z = 3, right end of 2.2+0.1 region), we exclude the two 0.1 regions. So sum all except the two 0.1s: 2.2+13.6+34.1+34.1+13.6+2.2. Let's compute: 2.2+2.2 = 4.4; 13.6+13.6 = 27.2; 34.1+34.1 = 68.2; 4.4+27.2 = 31.6; 31.6+68.2=99.8\approx99.7\% (due to empirical rule approximation). But looking at the student's note: 0.1+2.2 + 13.6+34.1+34.1+13.6+2.2+0.1 = 100, so between 67 and 85, we have 100 - 0.1 - 0.1=99.8\%, approximately 99.7\% (empirical rule for 3 standard deviations). But maybe the graph's percentages are: 0.1, 2.2, 13.6, 34.1, 34.1, 13.6, 2.2, 0.1. So adding the percentages between 67 and 85: 2.2 (67 - 70)+13.6 (70 - 73)+34.1 (73 - 76)+34.1 (76 - 79)+13.6 (79 - 82)+2.2 (82 - 85)=2.2+13.6=15.8; 15.8+34.1 = 49.9; 49.9+34.1=84; 84+13.6 = 97.6; 97.6+2.2=99.8\%.

Step1: Find the z - score for 79

$\mu = 76$, $\sigma = 3$, $x = 79$. $z=\frac{79 - 76}{3}=\frac{3}{3}=1$.

Step2: Use the normal distribution percentages

From the graph, the percentage of data less than or equal to $z = 1$ (79) is the sum of the left - hand regions: 0.1 (left of 67)+2.2 (67 - 70)+13.6 (70 - 73)+34.1 (73 - 76)+34.1 (76 - 79). Let's calculate: 0.1+2.2 = 2.3; 2.3+13.6 = 15.9; 15.9+34.1 = 50; 50+34.1 = 84.1\%. Using the empirical rule, the percentage of data less than $\mu+\sigma$ (since $z = 1$) is $50\%+34.1\% = 84.1\%$.

Step1: Recall the empirical rule for 1 standard deviation

The empirical rule states that approximately $68.2\%$ of the data lies within 1 standard deviation of the mean ($\mu-\sigma$ to $\mu+\sigma$).

Step2: Calculate the number of scores

The total number of scores is $n = 125$. The number of scores within 1 standard deviation is $0.682\times125$. Let's calculate: $0.682\times125=\frac{682}{1000}\times125=\frac{682}{8}=85.25\approx85$. (Alternatively, from the graph, the percentage within 1 standard deviation (73 - 79) is $34.1+34.1 = 68.2\%$. So $0.682\times125 = 85.25\approx85$)

Answer:

Approximately $99.7\%$ (or $99.8\%$ based on the graph)

Question 7b