Question

This Question: 1 pt 6 of 6 (0 complete) The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm and a standard deviation of 7.8 a. Find the probability that an individual distance is greater than 206.80 cm. b. Find the probability that the mean for 25 randomly selected distances is greater than 196.20 cm. c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30? a. The probability is (Round to four decimal places as needed.) b. The probability is (Round to four decimal places as needed.) Click to select your answer(s) * Previous

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Answer 1

Solution:- Given mean = 197.5, standard deviation = 7.8

a. P(X > 206.80) = P((X-μ)/σ > (206.80-197.5)/7.8)
= P(Z > 1.1923)
= 0.1170

b. For n = 25

P(X > 196.20) = P((X-μ)/(σ/sqrt(n)) > (196.20-197.5)/(7.8/sqrt(25)))
= P(Z > -0.8333)
= 0.7967

c. option C. The normal distribution can be used because the orginal population has a normal distribution.
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2. Z = 0.2533

x = μ + Z*σ = 100 + (0.2533*15) = 103.7995 = 103.8 (rounded)

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3. P(X < 129) = P(Z < (129-105)/20)
= P(Z < 1.2)
= 0.8849

The probability that a randomly selected adult has an IQ less than 129 is 0.8849

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4.

=> P(80 < X < 115) = P((80-100)/15 < Z < (115-100)/15))
= P(-1.3333 < Z < 1)
= 0.7495

The area of the shaded region is 0.7495