Question

The height, in inches, of a randomly chosen American woman is a normal random variable with mean μ = 64 and variance 2 = 7.84. (a) Calculate the probability that the height of a randomly chosen woman is between 59.8 and 68.2 inches. (b) Given that a randomly chosen woman is tall enough to be an astronaut (i.e., at least 59 inches tall), what is the conditional probability that she is at least 67 inches in height? (c) Four women are chosen at random. Calculate the probability that their total height is at least 260 inches. (d) Suppose that Z ∼ N(0, 1). Prove that E(Z19) = 0.

Answer 1

Calculations of normal distribution probabilities are based on converting the range values to z-scores, which allows us to find the likelihood of a woman's height falling within a certain range using the standard normal table.

Step-by-step explanation:

The scenario described involves a normal distribution, a very common continuous probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.

Given the heights of American women following a normal distribution with a mean of 64 inches and a variance of 7.84, we can calculate probabilities related to this distribution.

To calculate the probability of a woman's height falling within a certain range, we would use z-scores and the standard normal distribution table.

The z-score is a measure of how many standard deviations an element is from the mean. For a height of 59.8 inches and 68.2 inches, we first convert these to z-scores:

Z = (X - μ) / σ, where μ is the mean and σ is the standard deviation (which is the square root of the variance).

These calculations allow us to find the probability that a randomly chosen woman will have a height within the specified range.

Answer 2

Step-by-step explanation:

Given that the height in inches, of a randomly chosen American woman is a normal random variable with mean μ = 64 and variance 2 = 7.84.

X is N(64, 2.8)

Or Z =
`(x-64)/(2.8)`

a) the probability that the height of a randomly chosen woman is between 59.8 and 68.2 inches.

`=P(59.8<X<68.2)\\= P(|Z|<1.5)\\=0.8664`

b)
`P(X\geq 59)\\= P(X\geq -1.78)\\ \\=0.9625`

c) For 4 women to be height 260 inches is equivalent to

4x will be normal with mean (64*4) and std dev (2.8*4)

4x is N(266, 11.2)

`P(4x>260)= \\P(Z\geq -0.53571)\\=0.7054`

d) Z is N(0,1)

E(Z19) =
`P(Z>19)\\= 0.000`

since normal distribution is maximum only between 3 std deviations form the mean on either side.