Question
Solve the following questions 1.Mendel generated 580 offspring peas. He claimed that 75%,or 435, of them would have green pods. The actual experiment resulted in
Solve the following questions
1.Mendel generated 580 offspring peas. He claimed that 75%,or 435, of them would have green pods. The actual experiment resulted in 428 peas with green pods.
a)Assuming that groups of 580 offspring peas are generated, find the mean and standard deviation for the number of peas with green pods.
mean =n* p =580*75% = 435
std dev= (n*p*q) = (580*0.75*0.25() = 10.428
b)Use the range of thumb to find the minimum usual number and the maximum usual number of peas with green pods. Based on those numbers, can we conclude that Mendel's actual result of 428 peas with green pods is unusual? Does this suggest that Mendel's value of 75%is wrong?
minimum usual is 2 std deviations below the mean
minimum = 435 - 2*10.428 = 414.14
maximum usual is 2 std deviations above the mean
maximum = 435 + 2*10.428 = 455.86
2.Find the mean, Variance and Standard deviation for the probability distribution described in the following table.
0
0.001
1
0.015
2
0.088
3
0.396
4
0.237
5
0.237
3.If is a normally distributed variable with mean = 30 and standard deviation = 4. Find
a)
b)
c) P
4.A customer calling a call center spends an average of 45 minutes on hold during the peak season, with a standard deviation of 12 minutes. Suppose these times are normally distributed. Find the probability that the customer will be on hold for each interval of times:
a)More than 54 minutes
b)Less than 24 minutes.
c)Between 24 and 54 minutes.
5.Suppose from a random sample of size 49, we find a sample mean of $30. It is known that the typical distance a value is from the population (standard deviation) is $35. What is the population mean with 95% confidence?
6.Ten randomly selected automobiles were stopped, and the tread depth of the right front tire was measured. The mean was 0.32 inch, and the standard deviation was 0.08 inch. Find the 95% confidence interval of the mean depth. Assume that the variable is approximately normally distributed.
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