The mean number of minutes spent on a treadmill at the local gym is 28 minutes with a standard deviation of 8. If Kurt's z-score is 1.75, how much time did he spend on the treadmill?

To find out how much time Kurt spent on the treadmill, we need to use the formula for z-score:

z = (x - μ) / σ

where:
z = z-score
x = value of interest
μ = mean
σ = standard deviation

We are given that the mean (μ) is 28 minutes and the standard deviation (σ) is 8 minutes. We are also given that Kurt's z-score is 1.75. Let's plug these values into the formula and solve for x:

1.75 = (x - 28) / 8

Multiply both sides of the equation by 8 to isolate x:

1.75 * 8 = x - 28

14 = x - 28

Add 28 to both sides to isolate x:

14 + 28 = x

x = 42

Therefore, Kurt spent 42 minutes on the treadmill.

To find out how much time Kurt spent on the treadmill, we can use the formula for z-score:

z = (x - μ) / σ

where:
z is the z-score,
x is the raw score (the time Kurt spent on the treadmill),
μ is the mean,
and σ is the standard deviation.

In this case, we know the mean (μ) is 28 minutes and the z-score (z) is 1.75. The standard deviation (σ) is given as 8.

Let's rearrange the formula to solve for x:

x = (z * σ) + μ

Now, let's plug in the values:

x = (1.75 * 8) + 28
x = 14 + 28
x = 42

Therefore, Kurt spent 42 minutes on the treadmill.

well, he is 1.75 std above the mean, or

28 + 1.75 * 8