a normal distribution has a mean of 56 and a standard deviation of 8. Find the percentage of data values that are in the given interval. Use the curve to aid you.​

12) Between 40 and 64 = 0.815

13) Between 32 and 40 = 0.0235

14) Between 56 and 64 = 0.34

15) At most 56 = 0.515

16) At least 72 = 0.025

17) At most 64 = 0.855

Explanation:

To answer this, we will convert each of the values into their standardized form to make this easier.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ

x = Each value

μ = Mean = 56

σ = Standard deviation = 8

12) Between 40 and 64

For 40,

z = (x - μ)/σ = (40 - 56)/8 = (-16/8) = -2

For 64

z = (x - μ)/σ = (64 - 56)/8 = (8/8) = 1

So,

Between 40 and 64 = Between -2 and 1

From the curve, noting that the central point is the mean, with standard score of 0, the lines before it move in step of 1 standard deviation towards the negative side, that is, -1, -2, etc. And the lines before the central point move towards the positive side, that is, 1, 2, 3, etc.

So,

Between -2 and 1 = 0.135 + 0.34 + 0.34 = 0.815

13) Between 32 and 40

For 32,

z = (x - μ)/σ = (32 - 56)/8 = (-24/8) = -3

For 40,

z = (x - μ)/σ = (40 - 56)/8 = (-16/8) = -2

So,

Between 32 and 40 = Between -3 and -2 = 0.0235

14) Between 56 and 64

For 56,

z = (x - μ)/σ = (56 - 56)/8 = (0/8) = 0

For 64,

z = (x - μ)/σ = (64 - 56)/8 = (8/8) = 1

Between 56 and 64 = Between 0 and 1 = 0.34

15) At most 56

For 56,

z = (x - μ)/σ = (56 - 56)/8 = (0/8) = 0

At most 56 = At most 0 = 0.0015 + 0.0235 + 0.15 + 0.34 = 0.515

All the regions before z = 0)

16) At least 72

For 72,

z = (x - μ)/σ = (72 - 56)/8 = (16/8) = 2

At least 72 = At least 2 = 0.0235 + 0.0015 = 0.025

(All the regions from z = 2 to the end)

17) At most 64

For 64,

z = (x - μ)/σ = (64 - 56)/8 = (8/8) = 1

At most 64 = At most 1 = 0.0015 + 0.0235 + 0.15 + 0.34 + 0.34 = 0.855

(All the regions before z = 1)

Hope this Helps!!!

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