-38 due to the fact the lower the number, the colder it is.
Answer:![{ \left[\begin{array}{ccc}45.5\\\\\end{array}\right] }{}\\\\\end{array}\right] } ----](https://tex.z-dn.net/?f=%7B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D45.5%5C%5C%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%7D%7B%7D%5C%5C%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%7D%20----)
Step-by-step explanation: Hey, Basically we are going to arrange the data in an ascending order and the median is the middle value. If the number of values is an even number, the median will be the average of the two middle numbers.
Hope this helps!
Answer:
Mean and Variance of the number of defective bulbs are 0.5 and 0.475 respectively.
Step-by-step explanation:
Consider the provided information,
Let X is the number of defective bulbs.
Ten light bulbs are randomly selected.
The likelihood that a light bulb is defective is 5%.
Therefore sample size is = n = 10
Probability of a defective bulb = p = 0.05.
Therefore, q = 1 - p = 1 - 0.05 = 0.95
Mean of binomial random variable: 
Therefore, 
Variance of binomial random variable: 
Therefore, 
Mean and Variance of the number of defective bulbs are 0.5 and 0.475 respectively.