1) What is the z -score of x = 5, if it is 1.3 standard deviations to the left o

Question

1) What is the z-score of x = 5, if it is 1.3 standard deviations to the left of the mean? (Enter an exact number as an integer, fraction, or decimal.)

z =

2) Suppose X ~ N(6, 1). What value of x has a z-score of 3.25? (Enter an exact number as an integer, fraction, or decimal.)

3) If the area to the left of x in a normal distribution is 0.175, what is the area to the right of x? (Enter an exact number as an integer, fraction, or decimal.)

4)The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.4 days and a standard deviation of 2.1 days.

What is the probability of spending more than 2 days in recovery? (Round your answer to four decimal places.)

5) The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of 4 minutes and a standard deviation of 2 minutes.

Seventy percent of the time, it takes more than how many minutes to find a parking space? (Round your answer to two decimal places.)

6) A social network provides a variety of statistics on its website that detail the growth and popularity of the site. On average, 24 percent of 18 to 34-year-olds check their social network profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent.

(a) Find the probability that the percent of 18 to 34-year-olds who check the social network before getting out of bed in the morning is at least 27. (Round your answer to four decimal places.)

(b) Find the 95th percentile, and express it in a sentence. (Round your answer to two decimal places.)

95% of 18 to 34-year-olds who check the social network before getting out of bed in the morning is _____%

Explanation / Answer

Q4.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 5.4
standard Deviation ( sd )= 2.1
P(X > 2) = (2-5.4)/2.1
= -3.4/2.1 = -1.619
= P ( Z >-1.619) From Standard Normal Table
= 0.9473
Q5.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 4
standard Deviation ( sd )= 2
P ( Z > x ) = 0.7
Value of z to the cumulative probability of 0.7 from normal table is -0.5244
P( x-u / (s.d) > x - 4/2) = 0.7
That is, ( x - 4/2) = -0.5244
--> x = -0.5244 * 2+4 = 2.9512
Q6.
a.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 24
standard Deviation ( sd )= 5
P(X < 27) = (27-24)/5
= 3/5= 0.6
= P ( Z <0.6) From Standard Normal Table
= 0.7257
P(X > = 27) = (1 - P(X < 27)
= 1 - 0.7257 = 0.2743
b.
P ( Z > x ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is 1.6449
P( x-u / (s.d) > x - 27/5) = 0.05
That is, ( x - 27/5) = 1.6449
--> x = 1.6449 * 5+27 = 35.2243

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