here are yesterdays high temperatures (in fahrenheit) in 12 u.s. cities. 49, 50, 59, 59, 62, 66, 68, 75, 78…

here are yesterdays high temperatures (in fahrenheit) in 12 u.s. cities. 49, 50, 59, 59, 62, 66, 68, 75, 78, 80, 81, 85. notice that the temperatures are ordered from least to greatest. give the five - number summary and the interquartile range for the data set. five - number summary minimum: lower quartile: median: upper quartile: maximum: interquartile range:

here are yesterdays high temperatures (in fahrenheit) in 12 u.s. cities. 49, 50, 59, 59, 62, 66, 68, 75, 78, 80, 81, 85. notice that the temperatures are ordered from least to greatest. give the five - number summary and the interquartile range for the data set. five - number summary minimum: lower quartile: median: upper quartile: maximum: interquartile range:

Answer

Explanation:

Step1: Find the minimum

The smallest value in the data - set is 49. $49$

Step2: Find the maximum

The largest value in the data - set is 85. $85$

Step3: Calculate the median

Since there are $n = 12$ data points, the median is the average of the 6th and 7th ordered values. $\frac{66 + 68}{2}=67$. $67$

Step4: Calculate the lower half

The lower half of the data set consists of the first 6 values: 49, 50, 59, 59, 62, 66. The median of this lower half (lower quartile $Q_1$) is the average of the 3rd and 4th values. $\frac{59+59}{2}=59$. $59$

Step5: Calculate the upper half

The upper half of the data set consists of the last 6 values: 68, 75, 78, 80, 81, 85. The median of this upper half (upper quartile $Q_3$) is the average of the 3rd and 4th values. $\frac{78 + 80}{2}=79$. $79$

Step6: Calculate the inter - quartile range

The inter - quartile range (IQR) is $Q_3−Q_1$. So, $IQR=79 - 59=20$. $20$

Answer:

Minimum: 49 Lower quartile: 59 Median: 67 Upper quartile: 79 Maximum: 85 Interquartile range: 20