Chapter 6 1.) The amount of money thata student at Harvard carries on his/her pe

Question

Chapter 6

1.) The amount of money thata student at Harvard carries on his/her person is normally distributed with a mean of $25 and a standard deviation of $15. What is the probability that a randomly selected sample of 25 Harvard students will have an average amount less than $22 on his/her person? (Give your answer as a decimal rounded to 4 decimal places)

2.) Women's heights are normally distributed with a mean of 64.5 inches and standard deviation of 2.3 inches. If a woman is selected at random, what will be the probability that her height will be between 65.5 and 66.5 inches? (give your answer as a decimal rounded to 4 places)

Part II
Women's heights are normally distributed with a mean of 64.5 inches and a standard deviation of 2.3 inches. If a random sample of 12 women are selected, what will be the probability that the average height for the sample will be between 65.5 and 66.5 inches? (give your answer as a decimal rounded to 4 places)

3.) AAA reports that the average time it takes to respond to a roadside emergency is 25 minutes, with a standard deviation of 4.5 minutes. For a randomly selected sample of 5 emergency response calls what is the probability that the average wait time is less than 22 minutes for a roadside emergency? (give your answer as a rounded 4 place decimal)

4.)

The average number of hours that a student spends per day working at a computer is 3.5 hours. This distribution is normally distributed with a standard deviation of 0.9 hours. What percentage of students spend more than 3.8 hours per day working at a computer? (express your answer as a decimal rounded to four decimal places)

Part II
The average number of hours that a student spends per day working at a computer is 3.5 hours with a standard deviation of 0.9 hours. What is the probability that a randomly selected sample of 12 students will spend an average of more than 3.8 hours per day working on a computer? (express your answer as a decimal rounded to four decimal places)

5.)

The DMV reports that the average age of a vehicle in Santa Clara is 9 years old (108 months). Assume that the distribution of vehicle ages is normally distributed with a standard deviation of 18 months. What percent of vehicles in Santa Clara are between 10 (120 months) and 15 years old (180 months). (Express your answer as a decimal rounded to four decimal places)

Part II
The DMV reports that the average age of a vehicle in Santa Clara is 9 years old (108 months) with a standard deviation of 18 months. For a random sample of 20 vehicles in Santa Clara what is the probability that the average age of the vehicles in the sample is between 10 (120 months) and 15 years old (180 months). Give your answer as a decimal rounded to the fourth decimal place.

Explanation / Answer

1.

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    22      
u = mean =    25      
n = sample size =    25      
s = standard deviation =    15      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -1      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1   ) =    0.158655254 [ANSWER]

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2.

I.

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    65.5      
x2 = upper bound =    66.5      
u = mean =    64.5      
          
s = standard deviation =    2.3      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    0.434782609      
z2 = upper z score = (x2 - u) / s =    0.869565217      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.668139885      
P(z < z2) =    0.807730974      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.139591088   [ANSWER]

II.

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    65.5      
x2 = upper bound =    66.5      
u = mean =    64.5      
n = sample size =    12      
s = standard deviation =    2.3      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    1.506131137      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    3.012262274      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.933983244      
P(z < z2) =    0.998703458      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.064720214   [ANSWER]  
  

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