. For a normal distribution, find the percentage of data that are a. Within 1 standard deviation of the mean. b. To the right of 1.5 standard deviations below the mean c. More than 0.5 standard deviations away from the mean

a. The percentage of data within 1 standard deviation of the mean is 68.26%.

b. The percentage of data to the right of 1.5 standard deviations below the mean is of 93.32%.

c. The percentage of data more than 0.5 standard deviations away from the mean is of 61.7%.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

a. Within 1 standard deviation of the mean

Between and

This is the pvalue of Z = 1 subtracted by the pvalue of Z = -1.

Z = 1 has a pvalue of 0.8413

Z = -1 has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

The percentage of data within 1 standard deviation of the mean is 68.26%.

b. To the right of 1.5 standard deviations below the mean

Greater than , that is, 1 subtracted by the pvalue of Z = -1.5.

Z = -1.5 has a pvalue of 0.0668

1 - 0.0668 = 0.9332

The percentage of data to the right of 1.5 standard deviations below the mean is of 93.32%.

c. More than 0.5 standard deviations away from the mean

Below or above . These percentages are the same, so we find one and multiply by 1.

Below Z = -0.5

This is the pvalue of Z = -0.5, which is 0.3085

0.3085*2 = 0.617

The percentage of data more than 0.5 standard deviations away from the mean is of 61.7%.

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this