Question

1. For a standard normal distribution, determine the z-score, z₀, such that P(z < z₀) = 0.9698.

2. From a normal distribution with μμ = 76 and σσ = 5.9, samples of size 46 are chosen to create a sampling distribution. In the sampling distribution determine P(74.399 < ¯xx¯ < 78.079).

3. From a normal distribution with μμ = 81 and σσ = 2.7, samples of size 48 are chosen to create a sampling distribution. In the sampling distribution determine the probability that a sample mean is between 81.109 and 81.398.

4. From a normal distribution with μμ = 47 and σσ = 5.8, samples of size 33 are chosen to create a sampling distribution. In the sampling distribution determine the probability that a sample mean is at least 45.768.

5. From a normal distribution with μμ = 34 and σσ = 3.7, samples of size 46 are chosen to create a sampling distribution. In the sampling distribution determine the probability that a sample mean is at most 35.107.

Answer 1

1. For a standard normal distribution, determine the z-score, z₀, such that P(z < z₀) = 0.9698.

P(Z < 1.878) = 0.9698

2. From a normal distribution with μμ = 76 and σσ = 5.9, samples of size 46 are chosen to create a sampling distribution. In the sampling distribution determine P(74.399 < ¯xx¯ < 78.079).

Z score normal distribution formula:

z = (x - μ) / (σ/sqrt(n))

P(z< 74.399) = (74.399-76)/(5.9/sqrt(46)) = -1.84

P(z< 78.079) = (78.079-76)/(5.9/sqrt(46)) = 2.39

P(-1.84 < z < 2.39) = 0.9587

3. From a normal distribution with μμ = 81 and σσ = 2.7, samples of size 48 are chosen to create a sampling distribution. In the sampling distribution determine the probability that a sample mean is between 81.109 and 81.398.

Z score normal distribution formula:

z = (x - μ) / (σ/sqrt(n))

P(z< 81.109) = (81.109-81)/(2.7/sqrt(48)) = 0.28

P(z< 81.398) = (81.398-81)/(2.7/sqrt(48)) = 1.02

P(0.28 < z < 1.02) = 0.2359

4. From a normal distribution with μμ = 47 and σσ = 5.8, samples of size 33 are chosen to create a sampling distribution. In the sampling distribution determine the probability that a sample mean is at least 45.768.

Z score normal distribution formula:

z = (x - μ) / (σ/sqrt(n))

z= (45.768-47)/(5.8/sqrt(33)) = -1.22

P(z < -1.22) = 0.1112

5. From a normal distribution with μμ = 34 and σσ = 3.7, samples of size 46 are chosen to create a sampling distribution. In the sampling distribution determine the probability that a sample mean is at most 35.107.

Z score normal distribution formula:

z = (x - μ) / (σ/sqrt(n))

z= (35.107-34)/(3.7/sqrt(46)) = 2.03

P(z > 2.03) = 0.0212