Question

The Christmas Bird Count (CBC) is an annual tradition in Lexington, Massachusetts. A group of volunteers counts the number of birds of different species over a 1-day period. Each year, there are approximately 30–35 hours of observation time split among multiple volunteers. The following counts were obtained for the Northern Cardinal (or cardinal, in brief) for the period 2005–2011.
Year Number

2005 76

2006 47

2007 63

2008 53

2009 62

2010 69

2011 62

5.126 What is the mean number of cardinal birds per year observed from 2005 to 2011?

5.127 What is the standard deviation (sd) of the number of cardinal birds observed?

5.128 What is the probability of observing at least 60 cardinal birds in 2012? (Hint: Apply a continuity correction where appropriate.)

The observers wish to identify a normal range for the number of cardinal birds observed per year. The normal range will be defined as the interval (L, U), where L is the largest integer ≤ 15th percentile and U is the smallest integer ≥ 85th percentile.
5.129 If we make the same assumptions as in Problem 5.128, then what is L? What is U?

5.130 What is the probability that the number of cardinal birds will be ≥ U at least once on Christmas day during the 10-year period 2012–2021? (Hint: Make the same assumptions as in Problem 5.128.)

Answer 1

Answer:

(5.216) mean = 61.71

(5.217) standard deviation = 8.88

(5.218) P(X>=60) = 0.9238

(5.129) L = 79, U = 69

(5.130) P(X> or = U) = 0.7058

Step-by-step explanation:

The table of the statistic is set up as shown in attachment.

(5.216) mean = summation of all X ÷ no of data.

mean = 432/7 = 61.71 birds

(5.217)Standard deviation = √ sum of the absolute value of difference of X from mean ÷ number of data

S = √ /X - mean/ ÷ 7

= √551.428/7

S = 8.88

(5.218) P (X> or = 60)

= P(Z> or =60 - 61.71/8.8 )

= P(Z>or= - 0.192)

= 1 - P(Z< or = 0.192)

= 1- 0.0762

= 0.9238

(5.219)the 15th percentile=15/100 × 7

15th percentile = 1.05

The value is the number in the first position and that is 79,

L= 79

85th percentile = 85/100 × 7 = 5.95

The value is the number in the 6th position, and that is 69

U = 69

5.130) P(X>or = 60)

= P(Z>or= 69 - 61.71/8.8)

= P(Z> or = 0.8208)

= 1 - P(Z< or = 0.8209)

= 1 - 0.2942

= 0.7058