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The value of $z$corresponding to ${10}^{\text{th}}$percentile is $-1.28$.

a) By locating $0.10$in the Standard Normal Probability Table, you can determine the $z-$score that corresponds to the $10th$percentile.

$z-$scores are calculated based on the values at the beginning of each row and at the top of each column.

For example, the area closest to $0.10$in the Standard Normal Probabilities table is $0.1003(\mathrm{z}=-1.28)$.

The corresponding $z-$score (approximately) would be -1.28.

According to the following graph, the $10th$percentile and corresponding$z$value are represented as follows:

Therefore, the value of $z$corresponding to $10$^th percentile is $-1.28$

It is given in the question that, find the value of $34$% of all observations are greater than $z$

b) We are supposed to find the value of $z$such that $34\%$of all observations are greater than $z$.

If $34\%$of observations are above $z$values, it means that there exists $66\%$of observations are below$z$.

This means that the area of $0.66$in the Standard Normal Probabilities table corresponds approximately to $0.6591(\mathrm{z}=0.41)$.

Thus, the $z-$score (approximately) that corresponds to a $0.66$area is $0.41$.

Below is the graph showing the results.

As a result, $34\%$of observations are above $z$of $0.41$.