Z scores are given and we find areas corresponding to Z. Use Table 4. Find areas

Question

Z scores are given and we find areas corresponding to Z. Use Table 4.

Find areas for the given Z scores (see Guidelines on p. 239, and ex. 3, 4, 5, 6, in section 5.1):
1. Z = - 1.30
2. area to the right of Z = - 1.30
3. Z = - 1.17
4. area to the right of Z = 1.23
5. area between Z= -0.45 and Z = 2.11
6. area between Z= -1.13 and Z = 2.03

Section 5.2 - Find Z scores then find areas corresponding to Z. These area are our probabilities.
Find Probabilities (see ex. 1, 2 on p. 246, 247). Find key words such as "less than" means an area to the left of Z; "between" means an area between 2 Z scores; and "more than" means area to the right of Z.

1. In a survey of a group of men, the heights in the 20-29 age group were normally distributed, with a mean of 70.2 inches and a standard deviation of 3.0 inches. A study participant is randomly selected. Complete parts (a) through (d) below.
a). Find the probability that a study participant has a height that is less than 69 inches.
b). Find the probability that a study participant has a height that is between 69 and 73 inches.
c). Find the probability that a study participant has a height that is more than 73 inches.
d). Identify any unusual events. Explain your reasoning.

Section 5.3 - Find Z scores. We work in reverse: areas are given and we find corresponding Z scores.

Find Z scores for the given areas (see ex. 1 and 2, on p.252 and 253):

1. find Z for area = .4364
2. find Z for area = .9916
3 find Z for area = .6700
4. find Z where area to the right of Z is .0233
5. find Z where area to the right of Z is .7190
6. find Z for area = .9900

PART B - Empirical Rule
The scores for all high school students taking the math section of the SAT in a particular year had a mean of 480 and a standard deviation of 95. The distribution of SAT scores is bell-shaped. What percentage of students scored between 385 and 575 on this SAT test?

Explanation / Answer

1) as single value , hence area =0

2)P(Z>-1.30) =1-P(Z<-1.30) =1-0.0968=0.9032

3) as single value , hence area =0

4)P(Z>1.23) =1-P(Z<1.23) =1-0.8907=0.1093

5)P(-0.45<Z<2.11) =0.9826-0.3264 =0.6562

6)P(-1.13<Z<2.03) =0.9788-0.1292 =0.8496

Section 5.2

probability that a study participant has a height that is less than 69 inches=P(X<69) =P(Z<(69-70.2)/3)

=P(Z<-0.4) =0.3446

probability that a study participant has a height that is between 69 and 73 inches

P(69<X<73) =P((69-70.2)/3<(73-70.2)/3)=P(-0.4<Z<0.9333) =0.8247-0.3446 =0.4801

probability that a study participant has a height that is more than 73 inches

P(X>73) =1-P(Z<(73-70.2)/3)=1-P(Z<0.9333) =1-0.4801=0.5199

d) as all events are with in +/- 3 std deviation there is no unusual events.

Section 5.3

1)z=-0.1601

2)z=2.391

3)z=0.4399

4)1.9899

5)-0.5799

6) 2.3263

PART B

as 385 and 575 are 1 std deviation away from mean , from empirical formula, 68.27% values lie in b/w

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