Question

Suppose that a category of world class runners are known to run a marathon (26 miles) in an average of 146 minutes with a standard deviation of 12minutes. Consider 49 of the races.

Let X = the average of the 49 races.

a)Give the distribution of X. (Round your standard deviation to two decimal places.)

X~___(___,___)

b)Find the probability that the average of the sample will be between 145 and 148 minutes in these 49 marathons. (Round your answer to four decimal places.)

_________

c) Find the 70th percentile for the average of these 49 marathons. (Round your answer to two decimal places.)

_____min

d) Find the median of the average running times.

____min

Answer 1

Solution :

Given that,

mean = = 146

standard deviation = = 12

a) n = 49

= = 146

= / n = 12 / 49 = 4.54

The distribution of is approximately normal  N ( 146, 4.54)

b) P(145 < < 148)  

= P[(145 - 146) / 4.54< ( - ) / < (148- 146) / 4.54)]

= P( - 0.22 < Z < 0.44 )

= P(Z < 0.44) - P(Z < - 0.22 )

Using z table,  

= 0.6700 - 0.4129

= 0.2571

c) Using standard normal table,

P(Z < z) = 70 %

= P(Z < z) = 0.70

= P(Z < 0.52) = 0.70

z = 0.52

Using z-score formula  

= z * +

= 0.52 * 4.54 + 146

= 148.36

d) Using standard normal table,

P(Z < z) = 50 %

= P(Z < z) = 0.50

= P(Z < 0) = 0.50

z = 0

Using z-score formula  

= z * +

= 0 * 4.54 + 146

= 146

median = 146