Ruben and Victor both track the number of miles they walk each day for 6 months. The data is normally distributed for each student. Ruben had a mean μR of 5 miles with a standard deviation σR=1.1.

Victor had a mean μV of 4.4 miles with a standard deviation σV=1.4.

What are the probabilities that Ruben walked more than 6.1 miles and that Victor walked less than 5.8 miles? Select the probabilities that apply.

16%

34%

68%

545

95%

Step-by-step explanation:

Let x be the random variable representing the number of miles that each person walked each day for 6 months. Since it is normally distributed and the population mean and population standard deviation are known, we would apply the formula,

z = (x - µ)/σ

Where

x = sample mean

µ = population mean

σ = standard deviation

For Rueben,

µ = 5

σ = 1.1

the probability that Rueben walked more than 6.1 miles is expressed as

P(x > 6.1) = 1 - P( x ≤ 6.1)

For x = 6.1,

z = (4 - 6.1)/1.1 = - 1.91

Looking at the normal distribution table, the probability corresponding to the z score is 0.02807

P(x > 6.1) = 1 - 0.02807 = 0.97193

P(x > 6.1) = 0.97 × 100 = 97%

For Victor,

µ = 4.4

σ = 1.4

the probability that Victor walked less than 5.8 miless is expressed as

P(x < 5.8)

For x = 5.8,

z = (5.8 - 4.4)/1.4 = 1

Looking at the normal distribution table, the probability corresponding to the z score is 0.8413

P(x < 5.8) = 0.84 = 84%

the answer is d. (1.2, -4.7) on edgn

hope this !