Question

40 120 24023 decid head varia 1284 105 132 32064 2160623 .. Determine frequencies and relative frequencies for the herved values of number of nonconforming trans- ducers in a batch What proportion of batches in the sample have a most five nonconforming transducers? What proportion have fewer than five? What proportion have at least five non- conforming units? c. Draw a histogram of the data using relative frequency on the vertical scale, and comment on its features. 18. In a study of author productivity ("Lotka's Test." Collection Mem. 1982: 111-118), a large number of authors were classified according to the number of articles they had pub- lished during a certain period. The results were presented in the accompanying frequency distribution: Number of papers 1 2 3 4 5 6 7 8 Frequency 784 204 127 50 33 28 1919 Number 9 10 11 12 13 14 15 16 17 Frequency 6 7 6 7 4 4 5 3 3

Answer 1

Answer:

Given frequency distribution is as follows: 12 3 4 5 6 7 8 9 784 204 127 50 33 28 19 19 6 1x 10 11 12 13 14 15 16 17 16 7 4 4 5 3

Histogram for the given frequency distribution is shown below Histogram Frequency 1 2 3 4 5 6 12 13 14 15 16 17 7 8 9 10 11 Number of papers The histogram is completely skewed to the night which is highly positive skewed

From the given frequency distribution we can identity, the total number of authors published is 1309. Probability that number of authors who published at least 5 papers is P(X 25) = 1 - P(X <5) =1-{P( X = 1)+ P(X = 2)+ P(X = 3) + P( X = 4) 784 204 127 50 1309 1309 1309 1309 1165 1309 144 1309 =0.11 Therefore, the proportion of authors who published at least 5 papers is 11%

Probability that number of authors who published at least 10 papers is P(X 210) - 0.03 Therefore, the proportion of authors who published at least 10 papers is 3%

Probability that number of authors who published more than 10 papers is P(X >10) - 1300 0.024 Therefore, the proportion of authors who published more than 10 papers is 2.4%

15' has no upper No. It is impossible to draw a histogram because for the class' bound.

Yes. We take the width of the class is 2. So the area (11) of the bar equal to the product of height and width.