Aussie Maths & Science Teachers: Save your time with SmarterEd
The discrete random variable `X` has the probability mass function
`text(Pr)(X = x) = {(kx), (k), (0):} qquad {:(x∈{1, 4, 6}), (x = 3), (text(otherwise)):}`
Concerts at the Mathsland Concert Hall begin `L` minutes after the scheduled starting time. `L` is a random variable that is normally distributed with a mean of 10 minutes and a standard deviation of four minutes.
If a concert begins more than 15 minutes after the scheduled starting time, the cleaner is given an extra payment of $200. If a concert begins up to 15 minutes after the scheduled starting time, the cleaner is given an extra payment of $100. If a concert begins at or before the scheduled starting time, there is no extra payment for the cleaner.
Let `C` be the random variable that represents the extra payment for the cleaner, in dollars.
The owners of the Mathsland Concert Hall decide to review their operation. They study information from 1000 concerts at other similar venues, collected as a simple random sample. The sample value for the number of concerts that start more than 15 minutes after the scheduled starting time is 43.
The owners of the Mathsland Concert Hall decide that concerts must not begin before the scheduled starting time. They also make changes to reduce the number of concerts that begin after the scheduled starting time. Following these changes, `M` is the random variable that represents the number of minutes after the scheduled starting time that concerts begin. The probability density function for `M` is
`qquad qquad f(x) = {(8/(x + 2)^3), (0):} qquad {:(x ≥ 0), (x < 0):}`
where `x` is the the time, in minutes, after the scheduled starting time.
`qquad text(late is not within the sample 95% confidence interval.)`
a. `L\ ~\ N (10, 4^2)`
| `text(Pr) (L < 0)` | `= P(z < –2.5)` |
| `= 0.0062` |
| b. `text(Pr) (L > 15)` | `= text(Pr) ( z > 1.25)` |
| `= 0.1056` |
c.i.
| ii. `E(C)` | `= 0.8882 xx 100 + 0.1056 xx 200` |
| `= 109.94` | |
| `= $110 \ \ text((nearest dollar))` |
| iii. `E(C^2)` | `= 100^2 xx 0.8882 + 200^2 xx 0.1056` |
| `= 13\ 106` |
| `text(s.d.) (C)` | `= sqrt(E(C^2) – [E(C)]^2)` |
| `= sqrt(13\ 106 – (109.94)^2)` | |
| `= sqrt(1019.1 …)` | |
| `= 31.92 …` | |
| `= $32 \ \ text((nearest dollar))` |
d.i. `overset^p = (43)/(1000) \ \ , \ n = 1000`
| `95% \ text(C. I.)` | `= overset^p ± 1.96 sqrt((overset^p(1 – overset^p))/(n))` |
| `= (0.030, 0.056)` |
ii. `text(The proportion of concerts that begin more than 15 minutes)`
`text(late is not within the sample 95% confidence interval.)`
| e. `E(M)` | `= int_0^∞ x ((8)/((x +2)^3))\ dx` |
| `= 2` |
| f.i. `text(Pr)(M > 15)` | `=int_0^∞ x ((8)/((x +2)^3))\ dx` |
| `= (4)/(289)` |
ii. `text(Pr)(M > 15) = (4)/(289) \ \ , \ \ text(Pr)(M ≤ 15) = (285)/(289)`
`:. \ text(Pr) text{(9 concerts}\ \ M ≤ 15 ,\ text{10th concert}\ \ M > 15)`
`= ((205)/(289))^9 xx (4)/(289)`
`= 0.0122 \ \ text((to 4 decimal places))`
iii. `text(Pr)(15 < M < 20)= int_15^20 (8)/((x + 2)^3)\ dx = (195)/(34\ 969)`
`text(Pr)(M > 15)= int_15^oo (8)/((x + 2)^3)\ dx = (4)/(289)`
| `text(Pr)(M < 20 | M>15)` | `= (text(Pr)(15 < M < 20))/(text(Pr)(M > 15))` |
| `= ((195)/(34\ 969))/((4)/(289))` | |
| `= (195)/(484)` | |
| `= 0.403 \ \ text((to 3 decimal places))` |
The Lorenz birdwing is the largest butterfly in Town A.
The probability density function that describes its life span, `X`, in weeks, is given by
`f(x) = {(4/625 (5x^3 - x^4), quad 0 <= x <= 5),(0, quad text(elsewhere)):}`
The wingspans of Lorenz birdwing butterflies in Town A are normally distributed with a mean of 14.1 cm and a standard deviation of 2.1 cm.
Find the greatest possible wingspan, in centimetres, for a very small Lorenz birdwing butterfly in Town A, correct to one decimal place. (1 mark)
Each year, a detailed study is conducted on a random sample of 36 Lorenz birdwing butterflies in Town A.
A Lorenz birdwing butterfly is considered to be very large if its wingspan is greater than 17.5 cm. The probability that the wingspan of any Lorenz birdwing butterfly in Town A is greater than 17.5 cm is 0.0527, correct to four decimal places.
Find the smallest value of `n`, where `n` is an integer. (2 marks)
In a particular sample of Lorenz birdwing butterflies from Town B, an approximate 95% confidence interval for the proportion of butterflies that are very large was calculated to be (0.0234, 0.0866), correct to four decimal places.
Determine the sample size used in the calculation of this confidence interval. (2 marks)
The time Jennifer spends on her homework each day varies, but she does some homework every day.
The continuous random variable `T`, which models the time, `t,` in minutes, that Jennifer spends each day on her homework, has a probability density function `f`, where
`f(t) = {{:(1/625 (t - 20)),(1/625 (70 - t)),(0):}qquad{:(20 <= t < 45),(45 <= t <= 70),(text(elsewhere)):}:}`
Let `p` be the probability that on any given day Jennifer spends more than `d` minutes on her homework.
Let `q` be the probability that on two or three days out of seven randomly chosen days she spends more than `d` minutes on her homework.
| a. | |
MARKER’S COMMENT: Many did not draw graph along `t`-axis between 0 and 20 and for `t>70`.
b. `text(Pr)(25 <= T <= 55)`
`= int_25^45 1/625 (t – 20)\ dt + int_45^55 1/625 (70 – t)\ dt`
`= 4/5`
c. `text(Pr)(T ≤ 25 | T ≤ 55)`
`=(text(Pr)(T <= 25))/(text(Pr)( T <= 55))`
`= (int_20^25 1/625(t – 20)\ dt)/(1 – int_55^70 1/625(70 – t)\ dt)`
`= (1/50)/(1 – 9/50)`
`= 1/41`
d. `text(Pr)(T ≥ a) = 0.7`♦ Mean mark part (d) 36%.
`=>\ text(Pr)(T <= a) = 0.3`
`text(Solve:)`
| `int_20^a 1/625(t – 20)\ dt` | `= 0.3quadtext(for)quada ∈ (20, 45)` |
`:. a == 39.3649`
e.i. `text(Let)\ X =\ text(Number of days Jenn studies more than 50 min)`
`X ~\ text(Bi) (7, 8/25)`
`text(Pr)(X >= 4) = 0.1534`
| e.ii. | `text(Pr)(X >= 2 | X >= 1)` | `= (text(Pr)(X >= 2))/(text(Pr)(X >= 1))` |
| `= (0.7113…)/(0.9327…)` | ||
| `= 0.7626\ \ text{(to 4 d.p.)}` |
f. `text(Let)\ Y =\ text(Number of days Jenn spends more than)\ d\ text(min)`
`Y ~\ text(Bi)(7,p)`♦ Mean mark part (f) 36%.
| `q` | `= text(Pr)(Y = 2) + text(Pr)(Y = 3)` |
| `= ((7),(2))p^2(1 – p)^5 + ((7),(3))p^3(1 – p)^4` | |
| `= 21p^2(1 – p)^5 + 35p^3(1 – p)^4` | |
| `=7p^2(1-p)^4[3(1-p)+5p]` | |
| `=7p^2(1-p)^4(2p+3)` |
g.i. `text(Solve)\ \ q′(p) = 0,`♦♦ Mean mark part (g)(i) 30%.
| `p` | `=0.35388…` |
| `=0.3539\ \ text{(to 4 d.p.)}` |
`:. q_text(max)= 0.5665\ \ text{(to 4 d.p.)}`
g.ii. `text(Pr)(T > d) = p= 0.35388…`♦♦♦ Mean mark part (g)(ii) 8%.
`text(Solve:)`
| `int_d^70 (1/625(70 – t))dt` | `= 0.35388… quadtext(for)quadd ∈ (45,70)` |
| `:. d` | `=48.967…` |
| `=49\ text(mins)` |
The function
`f(x) = {{:(k),(0):}{:(sin(pix)qquadtext(if)qquadx ∈ [0,1]),(qquadqquadqquadqquadquadtext(otherwise)):}`
is a probability density function for the continuous random variable `X`.
A continuous random variable, `X`, has a probability density function given by
`f(x) = {{:(1/5e^(−x/5),x >= 0),(0, x < 0):}`
The median of `X` is `m`.
Find `text(Pr)(X < 1 | X <= m)`. (2 marks)