Mary Seitz invested a certain amount of money in a savings account paying 2% simple interest per year. When she withdrew her mon
1 answer:
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(a) Correct answer is Approximately normal because n less than or equals 0.05 Upper N and np left parenthesis 1 minus p right parenthesis greater than or equals 10.
(b) The value of P (X ≥ 770) is 0.0143.
(c) The value of P (X ≤ 720) is 0.0708.
Let X = number of elements with a particular characteristic.
The variable p is defined as the population proportion of elements with the particular characteristic.
The value of p is:
p = 0.74.
A sample of size, n = 1000 is selected from a population with this characteristic.
(a)
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:
μ = p
The standard deviation of this sampling distribution of sample proportion is:
σ =
The sample selected is of size, n = 1000 > 30.
Thus, according to the central limit theorem the distribution of is Normal, i.e. .
p~ N(μ = 0.74, σ =0.0139)
Thus the correct option is (A).
(b) We need to compute the value of P (X ≥ 770).
Apply continuity correction:
P (X ≥ 770) = P (X > 770 + 0.50)
= P (X > 770.50)
Then,
p > 770.5/1000 = 0.7705
Compute the value of P( p > 0.7705) as follows:
P( p > 0.7705) = P(p -μ/σ > 0.7705 - 0.74/0.0139)
= P( Z > 2.19)
= 1 - P( Z< 2.19)
= 1 - 0.98574
= 0.01426
≈ 0.0143
Thus, the value of P (X ≥ 770) is 0.0143.
(c)
We need to compute the value of P (X ≤ 720).
Apply continuity correction:
P (X ≤ 720) = P (X < 720 - 0.50)
= P (X < 719.50)
Then
Compute the value of as follows:
P( p < 0.7195) = P(p -μ/σ > 0.7705 - 0.74/0.0139)
= P(Z < - 1.47)
= 1 - P(Z < 1.47)
= 1 - 0.92922
= 0.07078
≈ 0.0708
Thus, the value of P (X ≤ 720) is 0.0708.
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