Answer:
The answer is below
Step-by-step explanation:
From the table, the mean (μ) = 1390.75 and the standard deviation (σ) = 518.75
The confidence level (C) = 90% = 0.9
α = 1 - C = 1 - 0.9 = 0.1
α/2 = 0.1 / 2 = 0.05
The z score of α/2 (0.05) is the same as the z score of 0.45 (0.5 - 0.05) which is equal to 1.645.
The margin of error (E) is given as:
The confidence interval = μ ± E = 1390.75 ± 228.07 = (1162.68, 1618.82)
The confidence interval is between 1162.68 and 1618.82.
Answer:
2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
The first step to solve this question is finding the proportion of students which use the computer more than 40 minutes, which is 1 subtracted by the pvalue of Z when X = 40. So
has a pvalue of 0.7881.
1 - 0.7881 = 0.2119
So 21.19% of the students use the computer for longer than 40 minutes.
Out of 10000
0.2119*10000 = 2119
2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.