Question

1. Statista reported in 2014 that the mean number of Facebook friends for 18 to 24 year olds was 649. Assuming the distribution is normal with mean 649 friends and standard deviation 80 friends, what is the probability that a randomly selected 18 to 24 year old will have between 565 and 822 friends?

P(565 < X < 822) =

Enter your answer as a number accurate to 4 decimal places. *Note: all z-scores must be rounded to the nearest hundredth.

2. Suppose that a brand of lightbulb lasts on average 1958 hours with a standard deviation of 100 hours. Assume the life of the lightbulb is normally distributed. Calculate the probability that a particular bulb will last from 1940 to 2196 hours?

P(1940 < X < 2196) =

Enter your answer as a number accurate to 4 decimal places.
*Note: all z-scores must be rounded to the nearest hundredth.

3. Suppose a drive for a PGA Tour golfer is 319.9 yards with a standard deviation of 32.7 yards. Find the probability that a random drive will travel more than 266.3 yards.

P(X > 266.3) =
Enter your answer as a number accurate to 4 decimal places. *Note: all z-scores must be rounded to the nearest hundredth.

Answer 1

(1) It is given that mean = 649, standard deviation = 80

To find :- probability between 565 and 822

Using normalcdf(lower bound, upper bound, mean, standard deviation)

puting lower bound = 565, upper bound =822, mean = 649 , standard deviation = 80

this implies

= normalcdf(565, 822, 649, 80)

= 0.8379

(2) It is given that mean = 1958, standard deviation = 100

To find :- probability between 1940 and 2196

Using normalcdf(lower bound, upper bound, mean, standard deviation)

puting lower bound = 1940, upper bound =2196, mean = 1958 , standard deviation = 100

this implies

= normalcdf(1940,2196,1958,100)

= 0.5628

(3) It is given that mean = 319.9, standard deviation = 32.7

To find :- probability greater than 266.3

Using normalcdf(lower bound, upper bound, mean, standard deviation)

puting lower bound = 266.3, upper bound =infinity or E99, mean = 319.9 , standard deviation = 32.7

this implies

= normalcdf(266.3,E99,319.9,32.7)

= 0.9494