Question

Find the area under the standard normal curve between z=−2.9 z = − 2.9 and z=0.28 z = 0.28 . Round your answer to four decimal places, if necessary.

Answer 1

The area under the standard normal curve between z = -2.9 and z = 0.28 is approximately 0.6161 when rounded to four decimal places.

Step-by-step explanation:

To find the area under the standard normal curve between z = -2.9 and z = 0.28, we use the cumulative distribution function (CDF) of the standard normal distribution. The CDF gives the probability that a standard normal random variable is less than or equal to a specific value.

First, we find the area to the left of z = -2.9 using the standard normal table or a calculator with a cumulative distribution function. The area to the left of z = -2.9 is approximately 0.0019.

Next, we find the area to the left of z = 0.28. This area is approximately 0.6103.

To find the area between z = -2.9 and z = 0.28, we subtract the area to the left of z = -2.9 from the area to the left of z = 0.28:

0.6103 - 0.0019 = 0.6161.

Therefore, the area under the standard normal curve between z = -2.9 and z = 0.28 is approximately 0.6161 when rounded to four decimal places. This represents the probability that a standard normal random variable falls between these two z-values on the standard normal distribution curve.

Answer 2

The area under the standard normal curve between z = -2.9 and z = 0.28 is approximately 0.6084.

Use a standard normal table or calculator:

These tools provide the cumulative area to the left of a given z-score.

Look up the cumulative areas:

Find the areas to the left of z = -2.9 and z = 0.28.

The area to the left of z = -2.9 is approximately 0.0019.

The area to the left of z = 0.28 is approximately 0.6103.

Subtract the areas:

Subtract the smaller area from the larger area to find the area between the two z-scores:

0.6103 (area to the left of 0.28) - 0.0019 (area to the left of -2.9) = 0.6084

Therefore, the area under the standard normal curve between z = -2.9 and z = 0.28 is approximately 0.6084.