Question

Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. (Round your answers to four decimal places.)(a) P(0 <= Z <= 2.24)(b) P(0 <= Z <= 2)(c) P(%u22122.60 <= Z <= 0)(d) P(%u22122.60 <= Z <= 2.60)(e) P(Z <= 1.64)(f) P(%u22121.75 <= Z)(g) P(%u22121.60 <= Z <= 2.00)(h) P(1.64 <= Z <= 2.50)(i) P(1.60 <= Z)(j) P(|Z| <= 2.50)

Answer 1

To calculate the probabilities for a standard normal random variable Z, one must find the area under the standard normal curve corresponding to the given Z values using a z-table or statistical software. Each scenario involves determining the area between, to the left, or to the right of specific Z values, considering the symmetric property of the normal distribution.

Step-by-step explanation:

Calculating Probabilities Using the Standard Normal Distribution

When working with a standard normal distribution, probabilities are found using areas under the curve. The standard normal distribution has a mean of 0 and a standard deviation of 1.

1. P(0 ≤ Z ≤ 2.24): This probability represents the area under the standard normal curve between 0 and 2.24. You would use a z-table or statistical software to find this value.
2. P(0 ≤ Z ≤ 2): Similar to the previous situation, this is the area between 0 and 2 under the curve.
3. P(-2.60 ≤ Z ≤ 0): This is the area under the curve from -2.60 to 0. It's the same as the area from 0 to 2.60 due to the symmetry of the distribution.
4. P(-2.60 ≤ Z ≤ 2.60): This calculation would encompass almost the entire area under the curve since -2.60 and 2.60 are far from the mean on both sides.
5. P(Z ≤ 1.64): The probability that Z is less than or equal to 1.64 is the area to the left of 1.64 under the curve.
6. P(-1.75 ≤ Z): This probability is the area under the curve to the right of -1.75, up to infinity.
7. P(-1.60 ≤ Z ≤ 2.00): Here, you find the area between -1.60 and 2.00.
8. P(1.64 ≤ Z ≤ 2.50): This represents the area under the curve between 1.64 and 2.50.
9. P(1.60 ≤ Z): The area to the right of 1.60 under the curve.
10. P(|Z| ≤ 2.50): This is the area between -2.50 and 2.50, which includes the central part of the distribution.

All probabilities found are between 0 and 1, and when considering the entire distribution, they sum to 1.

Answer 2

(a) P(0 <= Z <= 2.24) = 0.4927

(b) P(0 <= Z <= 2) = 0.4773

(c) P(-2.60 <= Z <= 0) = 0.4953

(d) P(-2.60 <= Z <= 2.60) = 0.9906

(e) P(Z <= 1.64) = 0.9495

(f) P(-1.75 <= Z) = 0.0047

(g) P(-1.60 <= Z <= 2.00) = 0.9425

(i) P(1.60 <= Z)=0.0548

(j) P(|Z| <= 2.50) = 0.9876

Step-by-step explanation:

(a) P(0 <= Z <= 2.24) = P(Z <= 2.24)- P(Z <= 0)

using the STANDARD NORMAL DISTRIBUTION TABLE

P(0 <= Z <= 2.24) = 0.9927 - 0.5 = 0.4927

(b) P(0 <= Z <= 2) = P(Z <= 2)- P(Z <= 0)

= 0.9773 - 0.5 = 0.4773

(c) P(-2.60 <= Z <= 0) = P(Z <= 0)- P(-2.60)

= 0.5 - 0.0047 = 0.4953

(d) P(-2.60 <= Z <= 2.60) = P(Z <= 2.6)- P(-2.60)

= 0.9953 - 0.0047 = 0.9906

(e) P(Z <= 1.64) = 0.9495

(f) P(-1.75 <= Z) =1 - P(Z < 2.6) = 1 - 0.9953 = 0.0047

(g) P(-1.60 <= Z <= 2.00) = P(Z <= 2.0)- P(-1.60)

= 0.9773- 0.0548 = 0.9425

(i) P(1.60 <= Z)=1 - P(Z < 1.6) = 1 - 0.9452 = 0.0548

(j) P(|Z| <= 2.50) = P(-2.5 < Z <= 2.50)= P(Z <= 2.5)- P(-2.5)

=0.9938 - 0.0062 = 0.9876