The average height of a woman aged 20 - 74 years is 64 inches in 2002, with an increase of approximately one inch from 1960. Suppose the height of a woman is normally distributed with a standard deviation of 2 inches. A) what is the probability that a randomly selected woman in this population is between 58 and 70 inches? B) what are the quartiles of this distribution? C) determine the height that is summetric about the mean that includes 90% of this population. D) what is the probability that 5 women selected at random from this population all exceed 68 inches?

sumasintihg

sumasintihg

Open question

2022-08-31

The average height of a woman aged 20 - 74 years is 64 inches in 2002, with an increase of approximately one inch from 1960. Suppose the height of a woman is normally distributed with a standard deviation of 2 inches.
A) what is the probability that a randomly selected woman in this population is between 58 and 70 inches?
B) what are the quartiles of this distribution?
C) determine the height that is summetric about the mean that includes 90% of this population.
D) what is the probability that 5 women selected at random from this population all exceed 68 inches?

Answer & Explanation

Marley Abbott

Marley Abbott

Beginner2022-09-01Added 4 answers

The average height of a woman aged 20 - 74 years is 64 inches in 2002, with an increase of approximately one inch from 1960. Suppose the height of a woman is normally distributed with a standard deviation of 2 inches.
A) what is the probability that a randomly selected woman in this population is between 58 and 70 inches? Z = 58 64 2 = 3 P ( Z < 3 ) = 0.0013 Z = 70 64 2 = 3 P ( Z < 3 ) = 0.9987 P ( 58 < X < 70 ) = P ( 3 < Z < 3 ) = P ( Z < 3 ) P ( Z < 3 ) = 0.9987 0.0013 = 0.99742
Answer: 0.9974
B) what are the quartiles of this distribution?
Q 1 : Z score corresponding to 25th percentile = -0.6745
0.6745 2 + 64 = 62.65
Q 1 = 62.65
Q 2 or median: Z score corresponding to 50th percentile = 0
0 2 + 64 = 64
Q 2 = 64
Q 3 : Z score corresponding to 75th percentile = 0.6745
0.6745 2 + 64 = 65.35
daufleguos

daufleguos

Beginner2022-09-02Added 6 answers

C) determine the height that is summetric about the mean that includes 90% of this population.
This questions sounds like you wanted a z score.
If you want the Z score, it is 1.64485 and -1.64485.
If you want the heights, this would be 3.2897 inches above and 3.2897 inches below, or60.71 and 67.29
D) what is the probability that 5 women selected at random from this population all exceed 68 inches?
Probability of exceeding 68 inches: Z score = 68 64 2 = 2.
P ( Z > 2 ) = 1 P ( Z < 2 ) = 1 0.9772 = 0.0228
Probability 5 random women exceeding 68 inches = 0.0228 5 = 0.00000000609

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