Answer:
And and if we use the following function on the Ti84 plus we got:
invNorm(0.02,0,1)
invNorm(1-0.02,0,1)
And the values with the middle 96% of the values are:
Step-by-step explanation:
For this case we want to find the limits with the middle 96% of the area below the normal curve, then the significance level would be:
And and if we use the following function on the Ti84 plus we got:
invNorm(0.02,0,1)
invNorm(1-0.02,0,1)
And the values with the middle 96% of the values are:
Answer:
A) attached below
B) 0.61
C) 0.47
Step-by-step explanation:
Given data:
Total number of lizards infected = 38
Of the 15 species B lizards 40% survived
For specie C one more survived than died
Out of the 24 lizards that died 1/3 were species A
<u>A) contingency table </u>
attached below
<u>B) Determine the proportion of these lizards in this study that were either specie A or Specie B </u>
P ( A or B ) = ( 8 + 15 ) / 38 = 0.605 ≈ 0.61
<u>C) determine the probability that specie C lizard did not study </u>
P ( not surviving | C ) = 7 / 15 = 0.466 ≈ 0.47
