Solve for n 10n+0.3=0.1+0.3
1 answer:
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Answer:
The probability that Polly's Sample will give a result within 1% of the value 65% is 0.6424
Step-by-step explanation:
The variable that assigns the value 1 if a person had its original major and 0 otherwise is a Bernoulli variable with paramenter 0.65. Since she asked the question to 1850 people, then the number of students that will have their original major is a Binomial random variable with parameters n = 1850, p = 0.65.
Since the sample is large enough, we can use the Central Limit Theorem to approximate that random variable to a Normal random variable, which we will denote X.
The parameters of X are determined with the mean and standard deviation of the Binomal that we are approximating. The mean is np = 1850*0.65 = 1202.5, and the standard deviation is √np(1-p) = √(1202.5*0.35) = 20.5152.
We want to know the probability that X is between 0.64*1850 = 1184 and 0.66*1850 = 1221 (that is, the percentage is between 64 and 66). In order to calculate this, we standarize X so that we can work with a standard normal random variable W ≈ N(0,1). The standarization is obtained by substracting the mean from X and dividing the result by the standard deviation, in other words
The values of the cummulative function of the standard normal variable W, which we will denote are tabulated and they can be found in the attached file.
Now, we are ready to compute the probability that X is between 1184 and 1221. Remember that, since the standard random variable is symmetric through 0, then \phi(-z) = 1-\phi(z) for each positive value z.
Therefore, the probability that Polly's Sample will give a result within 1% of the value 65% is 0.6424.
Answer:
a) 0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes
b) 0.0668 = 6.68% of the calls last more than 4.2 minutes
c) 0.0666 = 6.66% of the calls last between 4.2 and 5 minutes
d) 0.9330 = 93.30% of the calls last between 3 and 5 minutes
e) They last at least 4.3 minutes
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:
(a) What fraction of the calls last between 3.6 and 4.2 minutes?
This is the pvalue of Z when X = 4.2 subtracted by the pvalue of Z when X = 3.6.
X = 4.2
has a pvalue of 0.9332
X = 3.6
has a pvalue of 0.5
0.9332 - 0.5 = 0.4332
0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes
(b) What fraction of the calls last more than 4.2 minutes?
This is 1 subtracted by the pvalue of Z when X = 4.2. So
has a pvalue of 0.9332
1 - 0.9332 = 0.0668
0.0668 = 6.68% of the calls last more than 4.2 minutes
(c) What fraction of the calls last between 4.2 and 5 minutes?
This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 4.2. So
X = 5
has a pvalue of 0.9998
X = 4.2
has a pvalue of 0.9332
0.9998 - 0.9332 = 0.0666
0.0666 = 6.66% of the calls last between 4.2 and 5 minutes
(d) What fraction of the calls last between 3 and 5 minutes?
This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 3.
X = 5
has a pvalue of 0.9998
X = 3
has a pvalue of 0.0668
0.9998 - 0.0668 = 0.9330
0.9330 = 93.30% of the calls last between 3 and 5 minutes
(e) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4% of the calls. What is this time?
At least X minutes
X is the 100-4 = 96th percentile, which is found when Z has a pvalue of 0.96. So X when Z = 1.75.
They last at least 4.3 minutes