Answer:
a) If Y is distributed N(1, 4), Pr (Y ≤ 3) = 0.84134
b) If Y is distributed N(3, 9), Pr (Y > 0) = 0.84134
c) If Y is distributed N(50, 25), Pr (40 ≤ Y ≤ 52) = 0.63267
d) If Y is distributed N(5, 2), find Pr (6 ≤ Y ≤ 8) = 0.22185
Step-by-step explanation:
With the logical assumption that all of these distributions are normal distribution,
a) Y is distributed N(1, 4), find Pr ( Y ≤ 3 )
Mean = μ = 1
Standard deviation = √(variance) = √4 = 2
To find the required probability, we first standardize 3
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (y - μ)/σ = (3 - 1)/2 = 1
We'll use data from the normal probability table for these probabilities
The required probability
Pr ( Y ≤ 3 ) = P(z ≤ 1) = 0.84134
b) If Y is distributed N(3, 9), find Pr ( Y > 0 )
Mean = μ = 3
Standard deviation = √(variance) = √9 = 3
To find the required probability, we first standardize 0
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (y - μ)/σ = (0 - 3)/3 = -1
We'll use data from the normal probability table for these probabilities
The required probability
Pr ( Y > 0) = P(z > -1) = 1 - P(z ≤ -1) = 1 - 0.15866 = 0.84134
c) If Y is distributed N(50, 25), find Pr (40 ≤ Y ≤ 52).
Mean = μ = 50
Standard deviation = √(variance) = √25 = 5
To find the required probability, we first standardize 40 and 52.
For 40,
z = (y - μ)/σ = (40 - 50)/5 = -2
For 52,
z = (y - μ)/σ = (52 - 50)/5 = 0.4
We'll use data from the normal probability table for these probabilities
The required probability
Pr (40 ≤ Y ≤ 52) = P(-2.00 ≤ z ≤ 0.40)
= P(z ≤ 0.40) - P(z ≤ -2.00)
= 0.65542 - 0.02275
= 0.63267
d) If Y is distributed N(5, 2), find Pr ( 6 ≤ Y ≤ 8 )
Mean = μ = 5
Standard deviation = √(variance) = √2 = 1.414
To find the required probability, we first standardize 6 and 8.
For 6,
z = (y - μ)/σ = (6 - 5)/1.414 = 0.71
For 8,
z = (y - μ)/σ = (8 - 5)/1.414 = 2.12
We'll use data from the normal probability table for these probabilities
The required probability
Pr (6 ≤ Y ≤ 8) = P(0.71 ≤ z ≤ 2.12)
= P(z ≤ 2.12) - P(z ≤ 0.71)
= 0.983 - 0.76115
= 0.22185
Hope this Helps!!
Answer:
a) The value of N(1, 4) = 0.8413
b) The probability of N(3, 9) = 0.8413
ci) The probability (40≤ Y≤ 52) = 0.4
cii) The probability of N (3, 9) = 0.6236
d) The probability of (6≤Y≤8) = 0.2216
Step-by-step explanation:
Detailed step by step explanation is given in the attached document.
A normal distribution is a bell shaped symmetric distribution. This kind of distribution has a normal probability density function. A standard normal distribution is the one that has a mean 0 and variance of 1. It is often denoted as N (0, 1). If a general variance and mean are given and one has to look up probabilities in a normal probability distribution. The variable is standardized first. Standardizing a variable involves subtracting the general mean from the standard and then dividing the result by 1. In order to find the probabilities, the value of z is located in a normal distribution table.
