Using binomial distribution where success is the appearing of any of the top 10 most common names, thus probability of success (p) is 9.6% = 0.096 and the probability of failure = 1 - 0.096 = 0.904. Number of trials is 11.
Binomial distribution probability is given by P(x) = nCx (p)^x (q)^(n - x)
Probability that none of the top 10 most common names appears is P(0) = 11C0 (0.096)^0 (0.904)^(11 - 0) = (0.904)^11 = 0.3295
Thus, the probability that at least one of the 10 most common names appear is 1 - 0.3295 = 0.6705
Therefore, I will be supprised that none of the names of the authors were among the 10 most common names given that the probability that at least one of the names appear is 67%.
Answer:
The standard deviation of weight for this species of cockroaches is 4.62.
Step-by-step explanation:
Given : A different species of cockroach has weights that are approximately Normally distributed with a mean of 50 grams. After measuring the weights of many of these cockroaches, a lab assistant reports that 14% of the cockroaches weigh more than 55 grams.
To find : What is the approximate standard deviation of weight for this species of cockroaches?
Solution :
We have given,
Mean 
The sample mean x=55
A lab assistant reports that 14% of the cockroaches weigh more than 55 grams.
i.e. P(X>55)=14%=0.14
The total probability needs to sum up to 1,



The z-score value of 0.86 using z-score table is z=1.08.
Applying z-score formula,

Where,
is standard deviation
Substitute the values,





The standard deviation of weight for this species of cockroaches is 4.62.
Answer:
∠ e = 40°
Step-by-step explanation:
Since Δ ABC and Δ EDC are congruent then corresponding angles are congruent, that is
∠ A = ∠ E = 40°