Question

Reaction time is normally distributed, with a mean of 0.8 sec and a standard deviation of 0.1 sec. Find the probability that an individual selected at random has the following reaction times. (Round your answers to four decimal places.)

(a) greater than 1 sec


(b) less than 0.7 sec


(c) between 0.7 and 1 sec

Systolic blood pressure for a group of women is normally distributed, with a mean of 118 and a standard deviation of 9. Find the probability that a woman selected at random has the following blood pressures. (Round your answers to four decimal places.)

(a) greater than 133


(b) less than 111


(c) between 111 and 125

Answer 1

1.

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^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/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 0.8
standard Deviation ( sd )= 0.1/ Sqrt ( 1 ) =0.1
sample size (n) = 1
a.
the probability that an individual selected at random has the following reaction times greater than 1 sec
P(X > 1) = (1-0.8)/0.1/ Sqrt ( 1 )
= 0.2/0.1= 2
= P ( Z >2) From Standard Normal Table
= 0.0228
b.
the probability that an individual selected at random has the following reaction times less than 0.7 sec
P(X < 0.7) = (0.7-0.8)/0.1/ Sqrt ( 1 )
= -0.1/0.1= -1
= P ( Z <-1) From Standard NOrmal Table
= 0.1587
c.
the probability that an individual selected at random has the following reaction times between 0.7 and 1 sec
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 0.7) = (0.7-0.8)/0.1/ Sqrt ( 1 )
= -0.1/0.1
= -1
= P ( Z <-1) From Standard Normal Table
= 0.1587
P(X < 1) = (1-0.8)/0.1/ Sqrt ( 1 )
= 0.2/0.1 = 2
= P ( Z <2) From Standard Normal Table
= 0.9772
P(0.7 < X < 1) = 0.9772-0.1587 = 0.8186
2.
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^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 ) = 118
standard Deviation ( sd )= 9
a.
the probability that a woman selected at random has the following blood pressures greater than 133
P(X > 133) = (133-118)/9
= 15/9 = 1.6667
= P ( Z >1.6667) From Standard Normal Table
= 0.0478
b.
the probability that a woman selected at random has the following blood pressures less than 111
P(X < 111) = (111-118)/9
= -7/9= -0.7778
= P ( Z <-0.7778) From Standard Normal Table
= 0.2184
c.
the probability that a woman selected at random has the following blood pressures between 111 and 125
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 111) = (111-118)/9
= -7/9 = -0.7778
= P ( Z <-0.7778) From Standard Normal Table
= 0.2184
P(X < 125) = (125-118)/9
= 7/9 = 0.7778
= P ( Z <0.7778) From Standard Normal Table
= 0.7816
P(111 < X < 125) = 0.7816-0.2184 = 0.5633