Question

Given that z is a standard normal random variable, compute the following probabilities. a. P(−1.98≤z≤.49) b. P(.52≤z≤1.22) c. P(−1.75≤z≤−1.04)

Answer 1

a. P(-1.98 ≤ z ≤ 0.49) is approximately 0.6625.

b. P(0.52 ≤ z ≤ 1.22) is approximately 0.1901.

c. P(-1.75 ≤ z ≤ -1.04) is approximately 0.1091.

To compute these probabilities, we will use the standard normal distribution table or a calculator with a cumulative distribution function for the standard normal distribution (z-table). Here's how you can compute each of the probabilities step by step:

1. P(-1.98 ≤ z ≤ 0.49):

First, find the z-scores for -1.98 and 0.49 using the z-table or a calculator.

• For -1.98: The z-score is -1.98.
• For 0.49: The z-score is 0.49.

Now, use the cumulative distribution function (CDF) of the standard normal distribution to find the probabilities:

P(-1.98 ≤ z ≤ 0.49) = P(z ≤ 0.49) - P(z ≤ -1.98)

Look up the values in the z-table:

• P(z ≤ 0.49) ≈ 0.6864 (rounded to four decimal places)
• P(z ≤ -1.98) ≈ 0.0239 (rounded to four decimal places)

Now, subtract the two probabilities:

P(-1.98 ≤ z ≤ 0.49) ≈ 0.6864 - 0.0239 ≈ 0.6625 (rounded to four decimal places)

So, The answer is approximately 0.6625.

2. P(0.52 ≤ z ≤ 1.22):

Find the z-scores for 0.52 and 1.22:

• For 0.52: The z-score is 0.52.
• For 1.22: The z-score is 1.22.

Now, use the CDF of the standard normal distribution to find the probabilities:

P(0.52 ≤ z ≤ 1.22) = P(z ≤ 1.22) - P(z ≤ 0.52)

Look up the values in the z-table:

• P(z ≤ 1.22) ≈ 0.8892 (rounded to four decimal places)
• P(z ≤ 0.52) ≈ 0.6991 (rounded to four decimal places)

Now, subtract the two probabilities:

P(0.52 ≤ z ≤ 1.22) ≈ 0.8892 - 0.6991 ≈ 0.1901 (rounded to four decimal places)

So, The answer is approximately 0.1901.

3. P(-1.75 ≤ z ≤ -1.04):

Find the z-scores for -1.75 and -1.04:

• For -1.75: The z-score is -1.75.
• For -1.04: The z-score is -1.04.

Now, use the CDF of the standard normal distribution to find the probabilities:

P(-1.75 ≤ z ≤ -1.04) = P(z ≤ -1.04) - P(z ≤ -1.75)

Look up the values in the z-table:

• P(z ≤ -1.04) ≈ 0.1492 (rounded to four decimal places)
• P(z ≤ -1.75) ≈ 0.0401 (rounded to four decimal places)

Now, subtract the two probabilities:

P(-1.75 ≤ z ≤ -1.04) ≈ 0.1492 - 0.0401 ≈ 0.1091 (rounded to four decimal places)

So, The answer is approximately 0.1091.

Answer 2

Given that z is a standard normal random variable, compute the following probabilities;

a. P(-1.98 ≤ x ≤ 0.49) = 0.664.

b. P(-1.98 ≤ x ≤ 0.49) = 0.1903.

c. P(-1.75 ≤ x ≤ -1.04) = 0.1091.

In Mathematics and Statistics, the z-score of a given sample size or data set can be calculated by using the following formula:

Z-score, z = (x - μ)/σ

Where:

• σ represents the standard deviation.
• x represents the sample score.
• μ represents the mean score.

Part a.

Based on the standardized normal distribution table, the required probability for the given z-scores is given by:

P(-1.98 ≤ x ≤ 0.49) = P(z ≤ 0.49) - P(z ≤ -1.98)

P(-1.98 ≤ x ≤ 0.49) = 0.6879 - 0.0239

P(-1.98 ≤ x ≤ 0.49) = 0.664.

Part b.

P(0.52 ≤ x ≤ 1.22) = P(z ≤ 1.22) - P(z ≤ 0.52)

P(0.52 ≤ x ≤ 1.22) = 0.8888 - 0.6985

P(0.52 ≤ x ≤ 1.22) = 0.1903.

Part c.

P(-1.75 ≤ x ≤ -1.04) = P(z ≤ -1.04) - P(z ≤ -1.75)

P(-1.75 ≤ x ≤ -1.04) = 0.1492 - 0.0401

P(-1.75 ≤ x ≤ -1.04) = 0.1091.