5. Assume the distribution of minutes spent exercising per week i information th

Question

5. Assume the distribution of minutes spent exercising per week i information that the average amount of minutes spent exercising per deviation of 184 minutes. Suppose we take a sample of 200 people. s Normal. We know from past tandard week is 175 with a s a. What is the probability that a randomly selected person will exercise more than 420 minutes per week, averaging an hour per day? b. What is the probability that a randomly selected peron will xwercise les han 20 minutes per week, which equates to the 30 minutes per day suggested by healthcare professionals? hours per week? d. What is the probability that the average amount of exercise per week per person in our sample of 200 will be more than 210 minutes per week? (Meaning that on average, they exercise more than the recommended minimum amount). e, what amount of minutes defines the lowest 24.2% of the distribution for an individual person? f. What amount of minutes defines the highest 7.93% of the distribution for an individual person

Explanation / Answer

Question 5

average amount of minutes spent exercising per week = 175 minutes

standard deviation of minutes spent exercising per week = 175 minutes

(a) If x is the amount of time a random person spent exercising per week.

Pr(x > 420) = 1 - NORMAL (x < 420 ; 175 ; 184)

Z = (420 -175)/184 = 1.332

Pr(x > 420) = 1 - NORMAL (x < 420 ; 175 ; 184) = 1 - Pr(Z < 1.332) = 1 - 0.9086 = 0.0914

(b) Pr(x < 210 ; 175 ; 184)

Z = (210 - 175)/184 = 0.19

Pr(x < 210 ; 175 ; 184) = Pr(Z < 0.19) = 0.5753

(c) Here an hour (60 minutes) and 3 hours (180 minutes)

Pr(60 mins < x < 180 mins) = Pr(x < 180 mins) - Pr(x < 60 mins) = Pr(Z2) - Pr(Z1)

Z2 = (180 - 175)/184 = 0.0272; Z1 = (60 - 175)/184 = -0.625

Pr(60 mins < x < 180 mins) = Pr(x < 180 mins) - Pr(x < 60 mins) = Pr(Z2 < 0.0272) - Pr(Z1 < -0.625)

= 0.5109 - 0.2660 = 0.2449

(d) standard error of sample mean = /sqrt(n) = 184/ sqrt(200) = 13.01 minutes

Here lets say sample mean is x

so, Pr(x > 210 ; 175 ; 13.01) = 1 - Pr(x < 210 ; 175 ; 13.01)

Z = (210 - 175)/13.01 = 2.69

Pr(x > 210 ; 175 ; 13.01) = 1 - Pr(x < 210 ; 175 ; 13.01) = 1- Pr(Z > 2.69) = 1 - 0.9967 = 0.0033

(e) Here lets say the lowest 24.2% is k

so,

Pr(x < k ; 175 ; 13.01) = 0.242

Here z value form z table for the given probability value

Z = -0.7

so,

(k - 175)/184 = -0.7

k = 175 - 184 * 0.7 = 46.2 minutes

(f) let say the highest 7.93% if c

so,

Pr(x > c ; 175 mins ; 185 mins) = 0.0793

=> Pr(x < c ; 175 mins ; 185 mins) = 1 - 0.0793 = 0.9207

Here Z = 1.41 for the given p - value

(c - 175)/184 = 1.41

c = 175 + 184 * 1.41 = 434.44 minutes

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