Assume that women s heights are normally distributed with a mean given by μ=62.2 in, and a standard deviation given by σ=2.7 in.?
Update:
(a) If 1 woman is randomly selected, find the probability that her height is less than 63 in.
(b) If 38 women are randomly selected, find the probability that they have a mean height less than 63 in.
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a)
μ = 62.2
σ = 2.7
standardize x to z = (x - μ) / σ
P(x < 63) = P( z < (63-62.2) / 2.7)
= P(z < 0.2963) = 0.6179
(From Normal probability table)
b)
Mean μ = 62.2
Standard deviation σ = 2.7
Standard error σ / √ n = 2.7 / √ 38 = 0.4379978
standardize xbar to z = (xbar - μ) / (σ / √ n )
P(xbar < 63) = P( z < (63-62.2) / 0.437998)
= P(z < 1.8265) = 0.9664
(from normal probability table)
a) normalcdf(30,63,62.2,2.7)
.. = .6164980929 . . . . . . . . about 0.6165
b) normalcdf(30,63,62.2,2.7/√(38))
.. = .9661119855 . . . . . . . . about 0.9661
OK, here you go sweetheart :)
a) The mean is : μ = 62.2
The standard deviation is : σ = 2.7 /√1 = 2.7
Use the standardisation formula :
z = (X - μ) /σ
P(X < 63)
= P[z < (63 - 62.2)/2.7]
= P[z < 0.2963]
≈ 0.6165
Hence the required probability ≈ 61.65%
b) The mean is μ = 62.2
The standard deviation is : σ = 2.7 /√38 ≈ 0.438
Again, use the standardization formula :
z = (X - μ) /σ
P(X < 63)
= P[z < (63 - 62.2)/0.438]
= P[z < 1.826]
≈ 0.9661
Hence the required probability ≈ 96.61%
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