A survey of the mean number of cents off that coupons give was conducted by randomly surveying one coupon per page from the coup
on sections of a recent San Jose Mercury News. The following data were collected: 20% 75c; 50c; 65c; 300; 55 40C; 40C; 30C; 55, $1.50; 400; 65; 40C. Assume the underlying distribution is approximately normal. a. b. c. d. Define the random variables X and X in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population mean worth of coupons. i. State the confidence interval
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You multiply them just as you would multiply any normal polynomials, except there is a special rule. Since we know that i = √-1, the rule is that i² = -1. So, let's multiply:
Let's simplify this:
Therefore, your answer is 96 + 22i.
Answer:
0.7486 = 74.86% observations would be less than 5.79
Step-by-step explanation:
I suppose there was a small typing mistake, so i am going to use the distribution as N (5.43,0.54)
Problems of normally distributed samples can be 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.
The general format of the normal distribution is:
N(mean, standard deviation)
Which means that:
What proportion of observations would be less than 5.79?
This is the pvalue of Z when X = 5.79. So
has a pvalue of 0.7486
0.7486 = 74.86% observations would be less than 5.79