According to a study done by UCB students, the height for Martian adult males is normally distributed with an average of 67 inches and a standard deviation of 2.5 inches. Suppose one Martian adult male is randomly chosen. Let X = height of the individual. Round all answers to 4 decimal places where possible. %3D a. What is the distribution of X? X ~ N( 67 2.5 b. Find the probability that the person is between 63.7 and 64.9 inches. 0.1070

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On the planet of Mercury, 4-year-olds average 2.8 hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.3 hours and the amount of time spent alone is normally distributed. We randomly survey one Mercurian 4-year-old living in a rural area. We are interested in the amount of time X the child spends alone per day.

According to a study done by UCB students, the height for Martian adult males is normally distributed with
an average of 67 inches and a standard deviation of 2.5 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( 67
2.5
b. Find the probability that the person is between 63.7 and 64.9 inches. 0.1070
c. The middle 40% of Martian heights lie between what two numbers?
Low:
inches
High:
inches
Transcribed Image Text:According to a study done by UCB students, the height for Martian adult males is normally distributed with an average of 67 inches and a standard deviation of 2.5 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( 67 2.5 b. Find the probability that the person is between 63.7 and 64.9 inches. 0.1070 c. The middle 40% of Martian heights lie between what two numbers? Low: inches High: inches
Expert Solution
Step 1

Normal distribution: A continuous random variable x is said to follow normal distribution with parameter μ and σ2, if it's pdf is,

f(x)= 1σ2π×e-12σ2 (x-μ)2 ; -<x<, -<μ<, σ>0        = 0                                     ; otherwise

Here we calculate the probability by using z-score,

The formula is,

z = x-μσ

were,

x ~ normal distribution,

μ: Population mean.

σ: Population standard deviation.

Given,

Population mean = μ = 67

Population standard deviation = σ = 2.5

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