Stats

Eleven percent of students at a large high school are left-handed. A statistics teacher selects a random sample of 100 students and records X = the number of left-handed students in the sample. Calculate the mean of the sampling distribution of X.

(LT 6.3.1 #1)

• µ_X = 11

• µ_X = 3.13

• µ_X = 89

• µ_X = 100.11

Eleven percent of students at a large high school are left-handed. A statistics teacher selects a random sample of 100 students and records X = the number of left-handed students in the sample. Calculate the standard deviation of the sampling distribution of X.

(LT 6.3.1 #2)

• σ_X = 11

• σ_X = 3.13

• σ_X = 3.32

• σ_X = 10.01

Eleven percent of students at a large high school are left-handed. A statistics teacher selects a random sample of 100 students and records X = the number of left-handed students in the sample. Interpret the standard deviation of the sampling distribution of X, σ_X = 3.13.

(LT 6.3.1 #3)

• If many samples of size 100 were taken, the number of students who are left-handed would typically vary by about 3.13 from the mean of 11.

• If many samples of size 100 were taken, the number of students who are left-handed would typically vary by about 11 from the standard deviation of 3.13.

• If many samples of size 100 were taken, the number of students who are left-handed would typically vary by about 3.13 from the 11%.

• If many samples of size 100 were taken, the number of students who are left-handed would typically vary by about 11% from the standard deviation of 3.13.

Eleven percent of students at a large high school are left-handed. A statistics teacher selects a random sample of 100 students and records X = the number of left-handed students in the sample. Would it be appropriate to use a normal distribution to model the sampling distribution of X = the number of left-handed students in the sample? Justify your answer

(LT 6.3.2 #1)

• Yes, because np = 100(0.11) = 11 ≥ 10 and n(1-p) = 100(0.89) = 89 ≥ 10, the sampling distribution of X is approximately normal.

• Yes, because np = 100(0.11) = 11 < 10 and n(1-p) = 100(0.89) = 89 < 10, the sampling distribution of X is approximately normal.

• No, because np = 100(0.11) = 11 ≥ 10 and n(1-p) = 100(0.89) = 89 ≥ 10, the sampling distribution of X is not approximately normal.

• No, because np = 100(0.11) = 11 < 10 and n(1-p) = 100(0.89) = 89 < 10, the sampling distribution of X is not approximately normal.

Dysplasia is a malformation of the hip socket that is very common in certain dog breeds and causes arthritis as a dog gets older. According to the Orthopedic Foundation for Animals, 11.6% of all Labrador retrievers have hip dysplasia. A veterinarian tests a random sample of 50 Labrador retrievers and records Y = the number of Labs with dysplasia in the sample. Would it be appropriate to use a normal distribution of Y = the number of Labs with dysplasia in the sample? Justify your answer.

(LT 6.3.2 #2)

• No, because np = 50(0.116) = 5.8 < 10, the sampling distribution of Y is not approximately normal.

• No, because np = 50(0.116) = 5.8 > 10, the sampling distribution of Y is approximately normal.

• Yes, because np = 50(0.116) = 5.8 < 10, the sampling distribution of Y is not approximately normal.

• Yes, because np = 50(0.116) = 5.8 > 10, the sampling distribution of Y is approximately normal.

Calculate the probability that at least 15 of the members of the sample are left-handed with a sample size of 100 students and 11% of the students at a large high school are left-handed.

(LT 6.3.3 #1)

• 0.1006

• X is not approximately normal

• X is approximately normal and probability is 10.06

• X is not approximately normal and probability is 10.06

Calculate the probability that at most 5 of the individuals in the sample have never been married with a sample size of 50 and 20% of adults ages 25 and older have never been married.

(LT 6.3.3 #2)

• X is approximately normal and probability of 3.86

• X is not approximately normal and probability of 0.0386

• 0.0386

• X is not approximately normal and probability of 3.86