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

it's both, now can you answer my most recent question?

spinning a 1, 2, or 3. option d.

step-by-step explanation: when spinning a number graph up to 4, and aiming for any number less than 3, that automatically eliminates the number 4. now you have two numbers to spin on that are less than 3, but the number 3 is included in this question because it's no greater of an number than itself. i hope this you : )