In a study of exercise, a large group of male runners walk on a treadmill for six minutes. their heart rates in beats per minute at the end vary from runner to runner according to the n(104,12.5) distribution. the heart rates for male nonrunners after the same exercise have the n(130, 17) distribution.
(a) what percent of the runners have heart rates above 130?
(b) what percent of the nonrunners have heart rates above 130?

Using the normal probability relation, the probabilities that a runner and non - runner has a heart rate above 130 are 0.01876 and 0.5 respectively

Using the Zscore relation :

1.)

Runner with heart rate above 130 :

For X > 130 :

Zscore = (130 - 104) ÷ 12.5 = 2.08

P(Z > 2.08) = 1 - P(Z < 2.08)

Using a normal distribution table :

P(Z > 2.08) = 1 - 0.98124 = 0.01876

2.)

Non - Runner with heart rate above 130 :

For X > 130 :

Zscore = (130 - 130) ÷ 17 = 0

P(Z > 0) = 1 - P(Z < 0)

Using a normal distribution table :

P(Z > 0) = 1 - 0.5 = 0.5

Therefore, probability that a non - runner has a rate rate above 130 is 0.5

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a) 1.88% percent of the runners have heart rates above 130.

b) 50% percent of the nomrunners have heart rates above 130.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

(a) What percent of the runners have heart rates above 130?

In a study of exercise, a large group of male runners walk on a treadmill for six minutes. Their heart rates in beats per minute at the end vary from runner to runner according to the N(104,12.5) distribution. This means that .

This percentage is 1 subtracted by the pvalue of Z when .

has a pvalue of 0.9812.

This means that 1-0.9812 = 0.0188 = 1.88% percent of the runners have heart rates above 130.

(b) What percent of the nonrunners have heart rates above 130?

The heart rates for male nonrunners after the same exercise have the N(130, 17) distribution. This means that .

This percentage is 1 subtracted by the pvalue of Z when .

has a pvalue of 0.5000.

This means that 1-0.50 = 0.50 = 50% percent of the nomrunners have heart rates above 130.

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step-by-step explanation:

the total estimated nappy use.=10,500,000

step-by-step explanation:

given that on an average the average baby uses 3528 napkins.

in a particular year 2014, the number of births were 299697 in new south wales.

rounding off the number of births to the nearest hundred thousand we have

no of birtsh = 300000

also rounding off the average usage to nearst thousand we get

average use of a baby = 3500

hence total number of nappy estimated to be used = 3500x300000

= 10,500,000

answer is 10,500,000 nappies