A random variable X is normally distributed, with a mean of 34 and a standard de

Question

A random variable X is normally distributed, with a mean of 34 and a standard deviation of 5. Which of the following is the appropriate interquartile range for this distribution? The time it takes Alice to walk to the bus stop from her home is Normally distributed with mean 11 minutes and variance 3 minutes^2. The waiting time for the bus to arrive is Normally distributed with mean 5 minutes and standard deviation 1 minutes. Her bus journey to the UBC bus loop is a Normal variable with mean 24 and standard deviation 5 minutes. The time it take Alice to walk from the bus 9 loop to the lecture theatre to attend STAT 251 is Normally distributed with mean 18 minutes and variance 4 minutes^2. The total time taken for Alice to travel from her home to her STAT 251 lecture is Normally distributed. Please use R to find probabilities. What is the mean travel time (in minutes)? What is the standard deviation of Alice's travel time (in minutes, to 2 decimal places)? The STAT 251 class starts at 8 am sharp. Alice leaves home at 7 am. What is the probability (to 2 decimal places) that Alice will not be late for her class?

Explanation / Answer

1.

For Q1:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.25      
          
Then, using table or technology,          
          
z =    -0.67448975      
          
As x = u + z * s,          
          
where          
          
u = mean =    34      
z = the critical z score =    -0.67  
s = standard deviation =    5      
          
Then          
          
x = Q1 = 30.62      

For Q3:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.75      
          
Then, using table or technology,          
          
z =    0.67448975      
          
As x = u + z * s,          
          
where          
          
u = mean =    34      
z = the critical z score =    0.67      
s = standard deviation =    5      
          
Then          
          
x = critical value =    37.37244875 = 37.38

Hence,

IQR = 37.38-30.62 = 6.76 [ANSWER, C]  

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