The number of times a group of middle aged men have been to the gym in the past year is normal distributed with a mean of 33 times and a standard deviation of 8 times. Approximately what percentage (rounded to the nearest percent) of the men have been to the gym between 33 to 41 times?

P(33 ≤ x ≤ 41) = 0.34

Step-by-step explanation:

Since the number of times a group of middle aged men have been to the gym in the past year is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = number of times.

µ = mean

σ = standard deviation

From the information given,

µ = 33 times

σ = 8 times

The probability that the men have been to the gym between 33 to 41 times is expressed as

P(33 ≤ x ≤ 41)

For x = 33

z = (33 - 33)/8 = 0

Looking at the normal distribution table, the probability corresponding to the z score is 0.5

For x = 41

z = (41 - 33)/8 = 1

Looking at the normal distribution table, the probability corresponding to the z score is 0.84

P(33 ≤ x ≤ 41) = 0.84 - 0.5 = 0.34

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]