Answer :
Answer:
a) Sample mean = 100
Variance of the sampling distribution = 3.24
b) P(x > 102) = 0.1335
c) P(98 ≤ x ≤ 101) = 0.57876
d) P(x ≤ 101.5) = 0.79673
Step-by-step explanation:
For a given sample, the sample mean and standard deviation of the distribution of sample means are related to the mean and standard deviation of the population through
μₓ = μ = 100
σₓ = (σ/√n)
σ = population standard deviation = √(population variance) = √81 = 9
n = sample size = 25
σₓ = (σ/√n) = (9/√25) = 1.8
Variance of the sampling distribution = 1.8² = 3.24
b) Probability that the sample mean is greater than 102 P(x > 102)
This is a normal distribution problem
With mean = xbar = 100
And standard deviation = σ = 1.8
we need to first obtain the z-score of 102.
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - xbar)/σ = (102 - 100)/1.8 = 1.11
To determine the probability Probability that the sample mean is greater than 102
P(x > 102) = P(z > 1.11)
We'll use data from the normal probability table for these probabilities
P(x > 102) = P(z > 1.11) = 1 - P(z ≤ 1.11) = 1 - 0.86650 = 0.1335.
c) P(98 ≤ x ≤ 101)
We first normalize/standardize/obtain the z-scores of 98 and 101
For 98
z = (x - xbar)/σ = (98 - 100)/1.8 = -1.11
For 101
z = (x - xbar)/σ = (101 - 100)/1.8 = 0.56
P(98 ≤ x ≤ 101) = P(-1.11 ≤ z ≤ 0.56)
= P(z ≤ 0.56) - P(z ≤ -1.11)
= 0.71226 - 0.13350 = 0.57876
d) P(x ≤ 101.5)
Normalizing 101.5
z = (x - xbar)/σ = (101.5 - 100)/1.8 = 0.83
P(x ≤ 101.5) = P(z ≤ 0.83) = 0.79673
Hope this Helps!!!
