ID 26. Which of the following statements is TRUE about continuous probabilaty (a
ID: 3051290 • Letter: I
Question
ID 26. Which of the following statements is TRUE about continuous probabilaty (a) The uniform distribution has the lack of memory property (b) The uniform and normal distributions each have twability denst ro (c) The cumulative distribution function is the derivative o (e) The exponential distribution has its variance equal to normal distributions each have parameters greater thnted value 27. A recent study showed that the iPhone 6 battery life follows a normal diey of a randomly se the square root of its expected 24.5 hours and a standard deviation of 5.25 hours. What is the probabiing iPhone 6 lasting between 20 and 29 hours? distribution with a mean of of a randomly selected (a) 0.1949 (b) 0.8051 (c) 0.8571 (d) 0.6102 (e) 0.3898 28. Refer to the previous question. What is the battery life at the 90th percentile? (a) 17.780 (b) 29.225 (c) 28.783 (d) 31.220 (e) 33.136 29. Through various travels, an airplane inspector determines that, through a variety of factors, repai costs to an airplane in a major Canadian city follows a normal distribution with a mean of so thousand dollars and a standard deviation of 2 thousand dollars. Treating each city as independent, what is the probability that the total repair cost of one airplane from each of Edmonton, Winnipeg, Montréal, and Vancouver is more than 203,500 dollars? (a) 0.3300 (b) 0.8106 (c) 0.8750 (d) 0.6700 (e) 0.1894 30. Recently, Best Buy has experienced improved sales for BB-8 app-enabled droids, with 4 ou every 10 customers independently purchasing the item. What is the approximate probability no more than 90 out of 200 customers purchase the droid? (a) 0.4500 (b) 0.9357 (c) 0.5871 (d) 0.2188 (e) 0.0643Explanation / Answer
26) The uniform and normal distribution each have two parameters
27) mean= 24.5hours and standard deviation= 5.25 hours
Probability that it lasts between 20 and 29 hours
Since =24.5 and =5.25 we have:
P ( 20<X<29 )=P ( 2024.5< X<2924.5 )=P ((2024.5)/5.25<(X)/<(2924.5)/5.25)
Since Z=(x)/ , (2024.5)/5.25=0.86 and (2924.5)/5.25=0.86 we have:
P ( 20<X<29 )=P ( 0.86<Z<0.86 )
Use the standard normal table to conclude that:
P ( 0.86<Z<0.86 )=0.6102 (option d)
28) The battery life at the 90th percentile
Z score coressponding to 90th percentile is 1.29
Z= X-mu/sigma
1.29= X-24.5/5.25
1.29*5.25= X-24.5
6.7725=X-24.5
6.7725+24.5= X
31.2725 hours =X (option d)
29) mean= 50 thousand dollars and standard deviation= 2 thousand dollars
P(X>203500)
Since =50,000 and =2000 we have:
P ( X>203500 )=P ( X>203500050000 )=P ((X)/>(20350050000)/2000)
Since Z=(x)/ and (20350050000)/2000=76.75 we have:
P ( X>2035000 )=P ( Z>76.75 )
Use the standard normal table to conclude that:
P (Z>76.75)=0.1894
30) p= 0.4 n= 200 and x= 90
probability that no more than 90
by using excel function BINOMDIST i got 0.9357 ( option b)