Answer:
a) P(50 < x < 54) = 0.112
b) P(55 < x < 62) = 0.428
c) P(53 < x < 70) = 0.644
Step-by-step explanation:
Sample mean = nP = (sample size) × (population proportion) = 150 × 0.4 = 60
Sample standard deviation = √[nP(1-P)] = √(150×0.4×0.6) = 6
To check if this distribution approximates a normal distribution
np = 60 > 10
np(1-p) = 36 > 10
So, this is a normal distribution problem
a) P(50 < x < 54)
We standardize 50 and 54.
The standardized score for any value is the value minus the mean then divided by the standard deviation.
For 50
z = (x - μ)/σ = (50 - 60)/6 = - 1.67
For 54
z = (x - μ)/σ = (54 - 60)/6 = - 1.00
P(50 < x < 54) = P(-1.67 < z < -1.00) = P(z < -1.00) - P(z < -1.67) = 0.159 - 0.047 = 0.11
b) P(55 < x < 62)
We standardize 55 and 62.
The standardized score for any value is the value minus the mean then divided by the standard deviation.
For 55
z = (x - μ)/σ = (55 - 60)/6 = - 0.83
For 62
z = (x - μ)/σ = (54 - 60)/6 = 0.33
P(55 < x < 62) = P(-0.83 < z < 0.33) = P(z < 0.33) - P(z < -0.83) = 0.630 - 0.202 = 0.428
c) P(53 < x < 70)
P(53 < x < 70)
We standardize 53 and 70.
The standardized score for any value is the value minus the mean then divided by the standard deviation.
For 53
z = (x - μ)/σ = (53 - 60)/6 = - 0.50
For 70
z = (x - μ)/σ = (70 - 60)/6 = 1.67
P(53 < x < 70) = P(-0.50 < z < 1.67) = P(z < 1.67) - P(z < -0.50) = 0.953 - 0.309 = 0.644
Hope this Helps!!!