The following data are the joint temperatures of the O‐rings (°F) for each test firing or actual launch of the space shuttle rocket motor (from Presidential Commission on the Space Shuttle Challenger Accident, Vol. 1, pp. 129–131): 84, 49, 61, 40, 83, 67, 45, 66, 70, 69, 80, 58, 68, 60, 67, 72, 73, 70, 57, 63, 70, 78, 52, 67, 53, 67, 75, 61, 70, 81, 76, 79, 75, 76, 58, 31. Required:
a. Compute the sample mean and sample standard deviation.
b. Find the upper and lower quartiles of temperature.
c. Find the median.
d. Construct a box plot of the data and comment on the possible presence of outliers.

a

i) Sample mean = 65.861111

ii) Sample Standard deviation = 12.158836

b)

i) Upper quartile(Q3) = 75

ii) Lower quartile (Q1) = 58

c) Median = 67.5

d) We have an Outlier = 31

Step-by-step explanation:

84, 49, 61, 40, 83, 67, 45, 66, 70, 69, 80, 58, 68, 60, 67, 72, 73, 70, 57, 63, 70, 78, 52, 67, 53, 67, 75, 61, 70, 81, 76, 79, 75, 76, 58, 31.

Required:

a. Compute the sample mean and sample standard deviation.

Rearranging the data set

31, 40, 45, 49, 52, 53, 57, 58, 58, 60, 61, 61, 63, 66, 67, 67, 67, 67, 68, 69, 70, 70, 70, 70, 72, 73, 75, 75, 76, 76, 78, 79, 80, 81, 83, 84

i) Sample mean =

31 + 40 + 45 + 49 + 52 + 53 + 57+ 58 + 58 + 60 + 61 + 61 + 63 + 66 + 67 + 67 + 67 + 67 + 68 + 69 + 70 + 70,

+ 70 + 70 + 72 + 73 + 75 + 75 + 76 + 76 + 78 + 79 + 80 + 81 + 83 + 84/ 36

= 2371/36

= 65.861111

ii) Sample Standard deviation

= √(x - Mean)²/n - 1

= √[(31 - 65.861111)²+ (40 -65.861111)² +

(45- 65.861111)² +(49 -65.861111)² + (52 -65.861111)²+(53- 65.861111)² +(57 -65.861111)² +(58- 65.861111)² + (58 - 65.861111)² +(60 - 65.861111)²..................]/ 36 - 1

= √1215.297068 + 668.7970678+ 435.1859567 + 284.2970679 + 192.1304012 + 165.408179+ 78.51929012+ 61.7970679+ 61.7970679+ 34.35262346 + 23.63040123 + 23.63040123b+ 8.185956789 + 0.01929012346 + 1.297067901 + 1.297067901 + 1.297067901 + 1.297067901 + 4.574845679 + 9.852623457 + 17.13040124 + 17.13040124+ 17.13040124 + 17.13040124 + 37.68595679 + 50.96373457 + 83.51929013 + 83.51929013 + 102.7970679 + 102.7970679 + 147.3526235 + 172.6304013 + 199.908179 + 229.1859568 + 293.7415124 + 329.0192902/35

= √5174.305556/35

= √147.8373016

= 12.15883636

b. Find the upper and lower quartiles of temperature.

31, 40, 45, 49, 52, 53, 57, 58, 58, 60, 61, 61, 63, 66, 67, 67, 67, 67, 68, 69, 70, 70, 70, 70, 72, 73, 75, 75, 76, 76, 78, 79, 80, 81, 83, 84

i) Upper Quartile (Q3)

Formula = 3/4(n + 1)th value

n = 36

= 3/4(36 + 1)th value

= 111/4

= 27.75th value

Looking at the arranged value above

27.75th value = value between 27 and 28

= (27th + 28th)value/2

= 75 + 75/2

= 150/2

= 75

ii) Lower Quartile (Q1)

Formula = 1/4(n + 1)th value

n = 36

= 1/4(36 + 1)th value

= 37/4

= 9.25 th value

Looking at the arranged values above

9.25th value = value between 9 and 10

=(9th + 10th) value/2

58 + 60/2

= 118/2

= 59

c. Find the median.

31, 40, 45, 49, 52, 53, 57, 58, 58, 60, 61, 61, 63, 66, 67, 67, 67, 67, 68, 69, 70, 70, 70, 70, 72, 73, 75, 75, 76, 76, 78, 79, 80, 81, 83, 84

The above values have been arranged from ascending to descending order

The total number of values = 36

This is an even number.

Median formula = 1/2 (n + 1) the value

= 37/2

= 18.5th value

The median of this data set lies between = 18th and 19th value.

31, 40, 45, 49, 52, 53, 57, 58, 58, 60, 61, 61, 63, 66, 67, 67, 67,) 67, 68, (69, 70, 70, 70, 70, 72, 73, 75, 75, 76, 76, 78, 79, 80, 81, 83, 84

18th value = 67

19th value = 68

Median = 67 + 68/2

= 135/2

= 67.5

d. Construct a box plot of the data and comment on the possible presence of outliers.

Yes there's the possible presence of outliers.

This outlier = 31

Please find attached to this answer the diagram for the box plot.

3

Step-by-step explanation:

according to PEMDAS you divide before adding.

2/2 = 1

1 + 2 = 3!