For a recent 10k run, the finishers are normally distributed with mean 60 minute

Question

For a recent 10k run, the finishers are normally distributed with mean 60 minutes and standard deviation 14 minutes. Complete parts (a) through (d) below. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. a. Determine the percentage of finishers with times between 50 and 70 minutes. Approximately % of finishers had times between 50 and 70 minutes. Round to two decimal places as needed.) b. Determine the percentage of finishers with times less than 75 minutes. Approximately % of finishers had times less than 75 minutes Round to two decimal places as needed.) c. Obtain and interpret the 40th percentile for the finishing times. The 40th percentile isminutes. Round to two decimal places as needed.) Interpret the 40th percentile. Select the correct choice below and fill in the answer box(es) to complete your choice. Type integers or decimals rounded to two decimal places as needed.) A. This percentile is the value that is % greater than the smallest finishing time. O B. This percentile means that 96 of the finishing times are less than % of the finishing times are equal to O C. This percentile means that o. This percentile is the value that is % less than the average finishing time. d. Find the eighth decile for the finishing times The eighth decile is minutes. Round to two decimal places as needed.)

Explanation / Answer

a)P(50<X<70)=P((50-60)/14<Z<(70-60)/14)=P(-0.71<Z<0.71)=0.76.11%-23.89% =52.22%

b)

P(X<75)=P(Z<1.07)=85.77%

c)

for 40th percentile ; critical z =-0.25

hence 40th percentile =mean+z*std deviation=60-0.25*14=56.50

option B: this percentiel means that 40% of the finishing times are less than 56.50

d)

for eight decile ; critical z =0.84

corresponding value =mean+z*std deviaiton=60+0.84*14=71.76 minutes

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