Stat Exam 3

1. Use your z-table to find P(z < -1.43).

P(z < -1.43) = 0.0764

Use your z-table to find P(z > 2.95).

P(z > 2.95) = 1 - P(z � 2.95)
= 1 - 0.9984
= 0.0016

Use your z-table to find P(z � -3.01).

P(z � -3.01) = 1 - P(z < -3.01)
= 1 - 0.0013
= 0.9987

Use your z-table to find P(-0.67 < z < 1.08).

P(-0.67 < z < 1.08) = P(z < 1.08) - P(z � -0.67)
= 0.8599 - 0.2514
= 0.6085

Use your z-table to find P(-1.75 � z � 0.19).

P(-1.75 � z � 0.19) = P(z � 0.19) - P(z < -1.75)
= 0.5753 - 0.0401
= 0.5352

The lengths of human pregnancies are approximately normally distributed, with ? = 266 days and ? = 16 days. Show your work to find the probability that a randomly selected human pregnancy lasts no more than 245 days.

z =
?
x ?
=
16
245266
=
16
21
= 1.31
P(x � 245) = P(z � -1.31)
= 0.0951

The number of chocolate chips in an 18-ounce bag of Chips Ahoy chocolate chip cookies is approximately normally distributed, with ? = 1262 chips and ? = 118 chips. Show your work to find the probability that a randomly selected 18-ounce bag of Chips Ahoy

z =
?
x ?
=
118
15001262
=
118
238
= 2.02
P(x > 1500) = P(z > 2.02)
= 1 - P(z � 2.02)
= 1 - 0.9783
= 0.0217

The reading speed of 6th grade students is approximately normally distributed, with ? = 125 words per minute and ? = 24 words per minute. Show your work to find the probability that a randomly selected 6th grader reads between 100 and 165 words per minute

z1 =
?
x ? 1 =
24
100 125
=
24
25
= 1.04
z2 =
?
x ? 2 =
24
165125
=
24
40
= 1.67
P(100 < x < 165) = P(-1.04 < z < 1.67)
= P(z < 1.67) - P(z � -1.04)
= 0.9525 - 0.1492
= 0.8033

The amount of an ATM withdrawal at 1st Bank has an unknown distribution, with ? = $67 and ? = $35. If a random sample of 45 ATM withdrawals is obtained, show your work to find the probability that the average withdrawal is no less than $60. Be sure to exp

Since n = 45 (which is greater than 30), we can assume that the
x-distribution is approximately normal.
z =
n
?
x ?
=
45
35
60 67
=
5.2175
7
= 1.34
P(x � 60) = P(z � -1.34)
= 1 - P(z < -1.34)
= 1 - 0.0901
= 0.9099

The amount of time that adult Americans spend watching T.V. on a weekday is approximately normally distributed, with ? = 2.35 hours and ? = 1.93 hours. If a random sample of 24 adult Americans is obtained, show your work to find the probability that the a

Since the x-distribution is approximately normal, we can assume
that the x-distribution is also approximately normal.
z1 =
n
?
x ? 1 =
24
1.93
2.52.35
=
0.3940
0.15
= 0.38
z2 =
n
?
x ? 2 =
24
1.93
32.35
=
0.3940
0.65
= 1.65
P(2.5 � x � 3) = P(0.38 � z �