Temperature transducers of a certain type are shipped in batches of 50. A sample of 60 batches was selected, and the number of transducers in each batch not conforming to design specifications was determined, resulting in the following data. 5 1 2 4 0 1 3 2 0 5 3 3 1 3 2 4 7 0 2 3
3 4 2 1 3 1 1 3 4 1 2 3 2 2 8 4 5 1 3 1
8 0 2 3 2 1 0 6 4 2 1 6 0 3 3 3 6 1 2 3
(a) Determine frequencies and relative frequencies for the observed values of x = number of nonconforming transducers in a batch. (Enter relative frequencies to three decimal places.)
x Frequency Relati veFrequency
0
1
2
4
5
6
7
8
(b) What proportion of batches in the sample have at most six nonconforming transducers? (Enter your answer to three decimal places.)
What proportion have fewer than six? (Enter your answer to three decimal places.)
What proportion have at least six nonconforming units? (Enter your answer to three decimal places.)
(c) Draw a histogram of the data using relative frequency on the vertical scale, and comment on its features.
a) The center of the histogram is around 2 or 3 and it shows that there is some negative skewness in the data.
b) The center of the histogram is around 2 or 3 and it shows that the distribution is fairly symmetric.
c) The center of the histogram is around 4 or 5 and it shows that there is some positive skewness in the data.
d) The center of the histogram is around 2 or 3 and it shows that there is some positive skewness in the data.
e) The center of the histogram is around 4 or 5 and it shows that the distribution is fairly symmetric.
f) The center of the histogram is around 4 or 5 and it shows that there is some negative skewness in the data.

The number of six non conforming transducers at most =54/60=0.9

The number of fewer than six non conforming transducers =51/60=0.85

The number of at least six non conforming transducers at most =30/60=0.5

d) The center of the histogram is around 2 or 3 and it shows that there is some positive skewness in the data

Step-by-step explanation:

Relative FrequencyFrequency       x

6/60=0.1                            6                   0

0.2                                    12                    1

0.2                                   12                     2

0.25                            15                      3

0.1                                    6                       4

0.05                               3                 5

0.05                               3                 6

0.0167                              1                          7

0.033                               2                  8

∑                                    60                  

The number of six non conforming transducers at most

= 6+12+12+15+6+3= 54

The number of six non conforming transducers at most =54/60=0.9

The number of fewer than five non conforming transducers

= 6+12+12+15+6 = 51

The number of fewer than six non conforming transducers =51/60=0.85

The number of at least six non conforming transducers

2+1+3+3+6+15= 30

The number of at least six non conforming transducers at most =30/60=0.5

We plot the histogram by placing the percentages (or relative frequency *100) on the y –axis and the X on the x-axis.

The center of the histogram is given by the mean of the distribution which in this case is = ∑fx/∑f= 161/60= 2.683

So option d is the best answer

d) The center of the histogram is around 2 or 3 and it shows that there is some positive skewness in the data

Because at the center (between 2 and 3) the curve is increasing