Question 1199362: According to a study done by UCB students, the height for Martian adult males is normally distributed with an average of 64 inches and a standard deviation of 2.3 inches. Suppose one Martian adult male is randomly chosen. Let X = height of the individual. Round all answers to 4 decimal places where possible.
a. What is the distribution of X? X ~ N(64, 2.3)
b. Find the probability that the person is between 61.9 and 63.3 inches.
0.1998
c. The middle 20% of Martian heights lie between what two numbers?
Low:
inches
High:
inches
How do I solve letter C?
Answer by ikleyn(52790) (Show Source):
You can put this solution on YOUR website! .
Middle 20% are those who are lower than upper 60% and higher than lower 40%.
Use your calculator TI-83 or TI-84 with function invNorm.
Its format is p = invNorm ( probability, mu, SD).
For details and instructions about this function and how to use it, see web-site
https://www.statology.org/invnorm-ti-84/
The z-scores are
z_lower_40% = invNorm(0.4, 64, 2.3) = 63.4173.
z_upper_60% = invNorm(0.6, 64, 2.3) = 64.5827.
ANSWER. The middle 20% of Martian heights lie between 63.42 and 64.58 inches (after rounding).
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