Question

Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Round your answer to four decimal places.)
μ = 29;
σ = 3.4
P(x ≥ 30) = ______.

Answer 1

Answer:

P(x ≥ 30) = 0.3859

Step-by-step explanation:

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

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 question, we have that:

$\mu = 29, \sigma = 3.4$

This probability is going to be 1 subtracted by the pvalue of Z when X = 30. So

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

$Z = \frac{30 - 29}{3.4}$

$Z = 0.29$

$Z = 0.29$ has a pvalue of 0.6141

1 - 0.6141 = 0.3859

So

P(x ≥ 30) = 0.3859