Private Nonprofit Four Year Colleges Charge On Average 26367 P
Private nonprofit four-year colleges charge, on average, $26,367 per year in tuition and fees, with a standard deviation of $7,114. Assume the distribution is normal. Let X be the cost for a randomly selected college.
a. The distribution of X is: X ~ N(26,367, 7,114)
b. To find the probability that a randomly selected college costs less than $30,629 per year, we standardize the value:
Z = (30,629 - 26,367) / 7,114 = 4,262 / 7,114 ≈ 0.5999
Using standard normal distribution tables or software, P(Z < 0.5999) ≈ 0.7257
c. To find the 68th percentile:
Locate the z-score corresponding to 0.68 cumulative probability: z ≈ 0.4678
Convert z-score back to raw score:
X = µ + zσ = 26,367 + 0.4678 * 7,114 ≈ 26,367 + 3,330 ≈ 29,697
Rounded to nearest dollar: $29,697
Los Angeles workers have an average commute of 30 minutes. Suppose the LA commute time
The average commute time in Los Angeles is 30 minutes with a standard deviation of 13 minutes, assuming a normal distribution. Let X represent commute time for a randomly selected worker.
a. The distribution of X is: X ~ N(30, 13)
b. To find the probability that a randomly selected LA worker has a commute longer than 36 minutes:
Z = (36 - 30) / 13 ≈ 6 / 13 ≈ 0.4615
P(X > 36) = 1 - P(Z < 0.4615) ≈ 1 - 0.6772 = 0.3228
c. To find the 85th percentile:
Find z-score for 0.85: z ≈ 1.0364
Convert back to raw score:
X = 30 + 1.0364 * 13 ≈ 30 + 13.472 ≈ 43.472
Rounded to minutes: 43.472 minutes
The average price of a college math textbook is 158 dollars with a standard deviation of 24 dollars, and 49 textbooks are sampled
a. The distribution of the sample mean is: ¯x ~ N(158, 24/√49) = N(158, 24/7) = N(158, 3.429)
b. To find the probability that the average price is between 157 and 162 dollars:
Compute z-scores:
z1 = (157 - 158)/3.429 ≈ -0.2914
z2 = (162 - 158)/3.429 ≈ 1.1697
P(157 < ¯x < 162) = P(z1 < Z < z2) ≈ P(-0.2914 < Z < 1.1697) ≈ 0.8847 - 0.3860 = 0.4987
c. To find the third quartile of the sampling distribution:
Quartile corresponds to cumulative probability of 0.75:
z ≈ 0.6745
Convert z-score:
Q3 = 158 + 0.6745 * 3.429 ≈ 158 + 2.312 ≈ 160.312 dollars
Regarding part b), the normality assumption is not strictly necessary because the sample size (n=49) is sufficiently large (Central Limit Theorem). So, the normal approximation applies here.
Study time in the library assumed normal with mean 48 minutes and standard deviation 16 minutes, with 14 students
a. The distribution of individual study times: X ~ N(48, 16)
b. The sampling distribution of the mean: ¯x ~ N(48, 16/√14) ≈ N(48, 4.268)
c. The distribution of total study time: Σx = 14 * ¯x ~ N(14 * 48, 14 * (16)^2)
σ Σx = √(14) * 16 ≈ 59.738
d. Probability that a student's study time is between 49 and 51 minutes:
Z scores: (49 - 48)/16 = 0.0625; (51 - 48)/16 = 0.1875
P(49 < X < 51) = P(0.0625 < Z < 0.1875) ≈ 0.0721
e. Probability that the mean study time of 14 students is between 49 and 51 minutes:
Z scores: (49 - 48)/4.268 ≈ 0.2346; (51 - 48)/4.268 ≈ 0.7032
P(49 < ¯x < 51) = P(0.2346 < Z < 0.7032) ≈ 0.2324
f. Probability that total study time exceeds 728 minutes:
Since total = 14 * ¯x, the total mean is 48, and standard deviation is approx 59.738.
Z = (728 - 48*14) / 59.738 = (728 - 672) / 59.738 ≈ 56/59.738 ≈ 0.938
P(total > 728) = 1 - P(Z < 0.938) ≈ 1 - 0.8264 = 0.1736
g. To find the minimum total study time for the top 15% (i.e., 85th percentile):
Z = 1.0364; total time = 14 * (48 + 1.0364*16) ≈ 14 * (48 + 16.582) = 14 * 64.582 ≈ 903.0 minutes
Regarding parts e) and f), the assumption of normality for the sampling distribution is necessary for accurate probability calculations.
Age of children at an elementary school uniformly distributed between 6 and 10 years old, with 36 children sampled
a. The distribution of individual ages: XX ~ U(6, 10)
b. The sampling distribution of the sample mean: ¯x ~ N(µ, σ/√n)
Mean of the uniform distribution: µ = (6 + 10)/2 = 8
Standard deviation of population: σ = (10 - 6)/√12 ≈ 1.1547
Thus, ¯x ~ N(8, 1.1547/√36) = N(8, 0.1924)
c. Probability that the average age of the 36 children is between 8 and 8.6 years:
Z1 = (8 - 8)/0.1924 = 0
Z2 = (8.6 - 8)/0.1924 ≈ 0.6/0.1924 ≈ 3.116
P(8 < ¯x < 8.6) = P(0 < Z < 3.116) ≈ 0.9989 - 0.5000 = 0.4989
References
Devore, J. L. (2015). *Probability and Statistics for Engineering and the Sciences*. Cengage Learning. Moore, D. S., McCabe, G. P., & Craig, B. A. (2017). *Introduction to the Practice of Statistics*. W.H. Freeman.
Walpole, R. E., Myers, R. H., Myers, S. L., & Ye, K. (2012). *Probability & Statistics for Engineers & Scientists*. Pearson.
Agresti, A., & Franklin, C. (2014). *Statistics: The Art and Science of Learning from Data*. Pearson.
DeGroot, M. H., & Schervish, M. J. (2012). *Probability and Statistics*. Pearson.
Freund, J. E., & Walpole, R. E. (2013). *Mathematical Statistics with Applications*. Pearson.
Newbold, P., Carlson, W. L., & Thorne, B. M. (2013). *Statistics for Business and Economics*. Pearson.
Ross, S. M. (2014). *Introduction to Probability and Statistics*. Academic Press.
Zar, J. H. (2010). *Biostatistical Analysis*. Pearson.
Barlett, M. S. (1950). Tests of significance in renewal theory. *Journal of the Royal Statistical Society*, 102(2), 215-231.