Selected worked out answers for individual stats test.
3.
Vehicle Type Frequency
Motorcycle
8
Sedan
87
SUV
88
Truck
31
∑ 𝑥 = 214
8/214= 0.037, 87/214 = 0.407, 88/214 = 0.411, 31/214 = 0.145
4. 2,6 2,8, 4,6, and 4,8 are all the possible outcomes. 4 total outcomes.
Percentile 
 # of values below X   0.5 100%
6.
total # of values
3 values below 683, total number of value = 10, 3.5 / 10
7. IQR = Q3 – Q1 , Q3 = 74, Q1 = 65 IQR = 9
8.
CVAR 
s
100%
X
,
s = 3, x-bar = 25 3/25 = 0.12 => 12%
9. Use TI calculator. Place midpoints of each class of data in List1, place frequency count in L2, perform
1- Var Stats with L1 as x lists and L2 as frequency list. Answer is approximately 22.6
10. Tenth value (in ascending order) is 37.55, eleventh value is 37.85. (37.55 + 37.85)/2 = 37.70.
11. Use TI calculator. Place midpoints of each class of data in List1, place frequency count in L2, perform
1- Var Stats with L1 as x lists and L2 as frequency list. Answer is approximately 5.69
12. Sum frequencies of classes less than boundary of 28.5, 4 + 9 +12 = 25.
13. . Use TI calculator. Place value of X in List1, place probability of X in L2, and perform 1- Var Stats with
L1 as x lists and L2 as frequency list. Mean of probability distribution is 2.15.
14. Use TI calculator. Place value of X in List1, place probability of X in L2, and perform 1- Var Stats with
L1 as x lists and L2 as frequency list. Standard deviation is 1.307669683. Square this value to ascertain
variance. Variance = 1.71
15.
Loss/Gain
-20000
20000
35000
0
Probability
0.05
0.2
0.15
0.6
Value
-1000
4000
5250
0
Total = 8250
16. Use TI calculator. Place value of X (Visual response time) in List1, value of Y (Auditory response
time) in L2, and Linear regression calculation (LinReg a + bx). Value of a = 20.509336, value of b =
0.853345.
17. Use TI calculator. Place value of X in List1, value of Y in L2, and Linear regression calculation (LinReg
a + bx). Value r =0.814.
18. Use TI calculator Perform 1 Prop- Z test. Or use formula with p-hat value = .25. p = .31 and q = .69
z
and n = 88.
pˆ  p
pq n
Value of z = -1.22 (rounded to two decimal places for z score).
19. Use TI calculator T-Test. µ0 = 33700, X-bar = 36800, sx = 5000, N = 12, ergo d.f. = 11. T statistic =
2.148 (rounded to three decimal places).
20. Use TI calculator InvT function, area = 0.025 (one tail), d.f. = 15 use both positive and negative valus
for critical value of 2.131.
21. Use TI calculator Z-Test. µ0 = 60, X-bar = 64, σ = 23, n = 77, Z = 1.53, Probability z > 1.53 = 0.0635 Fail
To Reject H0
z 
ˆ ˆ  2 
n  pq
 E 
22.Input values in accordance with sample size formula,
2
P-hat = 0.38, q-hat = 0.62, Z critical value = 2.33, Margin of Error = 0.06 = 355.29 (always round up for
sample sizes to achieve minimum requirement).
23. Use TI calculator select 1-PropZInt (1 Proportion Z interval). X = 800*0.17 = 136, n = 800, C-Level =
.98 – Interval = (0.139, 0.201)
24. Use TI Calculator and select TInterval. X-bar = 8.7 sx = 1.1, n = 7, C-Level = .95. (7.7, 9.7).
25. Use TI Calculator and select ZInterval (population standard deviation is given). X-bar = 13.8 σ = 7.3, n
= 35, C-Level = .95. (11.4, 16.2)
26. Use TI calculator and select DISTR function. Select normalcdf (normal distribution with cumulative
density function). Lower limit = 2.6, upper limit = infinity (E99 on TI calculator), population mean = 3,
standard error of the mean = 1/ √9. Probability that the nine selected sand dollars have a diameter
greater than 2.6 cm is 0.885 (approximately).
27. Standard error of mean = σ/√𝑛, 55/√25 =
55
5
P X  
28. Problem is a binomial problem.
= 11.
n!
 p X  q n X
 n  X ! X !
P(3), n = 5, p = 0.25, q = 1-p = 0.75, x = 3, n-x = 2. P(3) = 0.088 approximately.
29. Probability X>10. Us TI calculator. Select DISTR, select normalcdf function. Input lower limit of 10,
upper limit of infinity, mean of 5.7 and standard deviation of 1.8. P(x > 10) = 0.008 or 0.8%.
30. . Probability X>999. Us TI calculator. Select DISTR, select normalcdf function. Input lower limit of 999,
upper limit of infinity, mean of 998 and standard deviation of 7. P(x > 10) = 0.4432.