question 1 elevan peas are generated from parents having the

question 1:

elevan peas are generated from parents having the green/yellow pair of genes, so there is a 0.75 probability that an individual pea will have a green pod. find the probability that among the 11 offspring peas, at least 10 have green pods. is it unusual to get at least 10 peas with green pods when 11 offspring peas are generated? why or why not? The probability that at least 10 of the 11 offspring peas have green pods is?

QUESTION 2.

Assume that the given procedure yields a binomial distribution with n trials and the probability of success for one trial is p. use the given values of n and p to find the mean and stard deviation. Also, use the range rule of thumb to find the minumum usual value u-2o and maximum usual value u +2o. in an analysis of prelimanry test results from the a gender-selection method, 31 babies are born and it is assumed that 50% of babies are girls, so n=31 and p=0.5. the value of the mean is?

question 3.

Data from 14 cities were combined for a 20-year period, and the total 280 city-year included a total of 140 homicides. after finding the mean number of homicides per city year, find the probability that a randomly selected city-year has the following numbers of homicides, then compare the actual results to those expected by using the Poisson probabilities. P(0) =

QUESTION 4.

The accompanying table describes results from eight offsprings. the random variable x represents the number of offspring peas with green pods.

A. FIND the probability of getting exactly 7 peas with green pods

QUESTION 5:

The accompanying table describes results of roadworthiness tests of cars that are 3 years old. The random variable x represents the numbers of cars that failed among six that were tested for roadworthiness. Find the mean and standard deviation for the number of cars that failed among the six cars that are tested. find the mean for the number of cars that failed.

Solution

Q1.

Binomial Distribution

PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial

P( X = 10 ) = ( 11 10 ) * ( 0.75^10) * ( 1 - 0.75 )^1
= 0.155
P( X = 11 ) = ( 11 11 ) * ( 0.75^11) * ( 1 - 0.75 )^0
= 0.042
P( X > = 10 ) = 0.155 + 0.042= 0.197

No, it is not unusual to get atleast 10 Green Pods, as the
reason probability of such 0.197 > 0.05

Q4.
probability of getting exactly 7 peas with green pods = 0.284