Question

1. The weight of 6-week-old poults (juvenile turkeys) are normally distributed with a mean 8.6 pounds and standard deviation 1.9 pounds. Find the first quartile of the weight.

A) 8.80 lb B) 8.06 lb C) 6.60 lb D) 7.33 lb

2. A gardener buys a package of seeds. Eighty-seven percent of seeds of this type germinate. The gardener plants 130 seeds. Approximate the probability that 81% or more seeds germinate.

A) 0.9761 B) 0.9884 C) 0.9904 D) 0.9842

3. A population has a standard deviation σ = 20.2. How large a sample must be drawn so that a 95% confidence interval for μ will have a margin of error equal to 4.2?

4. A college admissions officer takes a simple random sample of 120 entering freshmen and computes their mean mathematics SAT score to be 448. Assume the population standard deviation is 116. Based on a 98% confidence interval for the mean mathematics SAT score, is it likely that the mean mathematics SAT score for entering freshmen class is greater than 464? (Hint: you should first construct the confidence interval for the mean mathematics SAT score.). Answer Yes or No and give the reasoning for your answer.

Answer 1

Question 1

X ~ N ( µ = 8.6 , σ = 1.9 )
P ( X < x ) = 25% = 0.25
To find the value of x
Looking for the probability 0.25 in standard normal table to calculate Z score = -0.6745
Z = ( X - µ ) / σ
-0.6745 = ( X - 8.6 ) / 1.9
X = 7.33
P ( X < 7.33 ) = 0.25

D) 7.33 lb

Question 2

Condition check for Normal Approximation to Binomial
n * P >= 10 = 130 * 0.87 = 113.1
n * (1 - P ) >= 10 = 130 * ( 1 - 0.87 ) = 16.9

Using Normal Approximation to Binomial
Mean = n * P = ( 130 * 0.87 ) = 113.1
Variance = n * P * Q = ( 130 * 0.87 * 0.13 ) = 14.703
Standard deviation = √(variance) = √(14.703) = 3.8344

P ( X >= 105.3 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 105.3 - 0.5 ) =P ( X > 104.8 )

X ~ N ( µ = 113.1 , σ = 3.8344 )
P ( X > 104.8 ) = 1 - P ( X < 104.8 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 104.8 - 113.1 ) / 3.8344
Z = -2.16
P ( ( X - µ ) / σ ) > ( 104.8 - 113.1 ) / 3.8344 )
P ( Z > -2.16 )
P ( X > 104.8 ) = 1 - P ( Z < -2.16 )
P ( X > 104.8 ) = 1 - 0.0158
P ( X > 104.8 ) = 0.9842

D) 0.9842

Question 3

Sample size can be calculated by below formula
n = (( Z(α/2) * σ) / e )2
n = (( Z(0.05/2) * 20.2 ) / 4.2 )2
Critical value Z(α/2) = Z(0.05/2) = 1.96
n = (( 1.96 * 20.2 ) / 4.2 )2
n = 89
Required sample size at 95% confident is 89.

Question 4

Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.02 /2) = 2.326
448 ± Z (0.02/2 ) * 116/√(120)
Lower Limit = 448 - Z(0.02/2) 116/√(120)
Lower Limit = 423.3693
Upper Limit = 448 + Z(0.02/2) 116/√(120)
Upper Limit = 472.6307
98% Confidence interval is ( 423.3693 , 472.6307 )

No, Since value 464 lies in the interval, hence we fail to conclude that  the mean mathematics SAT score for entering freshmen class is greater than 464.