Question

A normal distribution has a mean of 137 and a standard deviation of 7. Find the z-score for a data value of 121. Incorrect Round to two decimal places g

Answer 1

Answer:

$-2.29$.

Step-by-step explanation:

We have been given that a normal distribution has a mean of 137 and a standard deviation of 7. We are asked to find the z-score corresponding to data value 121.

We will use z-score formula to solve our given problem.

$z=\frac{x-\mu}{\sigma}$, where,

z = Z-score,

x = Sample score,

$\mu$ = Mean,

$\sigma$ = Standard deviation.

Upon substituting our given values in above formula, we will get:

$z=\frac{121-137}{7}$

$z=\frac{-16}{7}$

$z=-2.2857142857$

Upon rounding to two decimal places, we will get:

$z\approx -2.29$

Therefore, z-score corresponding to data value 121 would be $-2.29$.

Answer 2

Answer:

The z-score for a data value of 121 is -2.29.

Step-by-step explanation:

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

$Z = \frac{X - \mu}{\sigma}$

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.

In this problem, we have that:

$\mu = 137, \sigma = 7$

Find the z-score for a data value of 121.

This is Z when X = 121. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{121 - 137}{7}$

$Z = -2.29$

The z-score for a data value of 121 is -2.29.