Normal Distribution in Probability
$begingroup$
I can solve the k, but I am not really able to do the rest!
The mass M of apples in grams is normally distributed with mean μ . The following table shows probabilities for values of M .
Values: M>93 -> p(x)=k
93< M < 119 -> p(x)= 0.98
119 < M -> p(x)= 0.01
Part a:
(i) Write down the value of k .
(ii) Show that μ = 106.
Part b:
Find P (M < 95) .
Part c and d:
The apples are packed in bags of ten.Any apples with a mass less than 95 g are classiied as small.
Find the probability that a bag of apples selected at random contains at most one small apple.
A crate contains 50 bags of apples. A crate is selected at random.
(i) Find the expected number of bags in this crate that contain at most one small apple.
(ii) Find the probability that at least 48 bags in this crate contain at most one small apple.
probability statistics
edited Dec 12 '18 at 13:24
David G. Stork
11k31432
asked Dec 12 '18 at 10:30
Dominik Dominik
83
$endgroup$
add a comment |
$begingroup$
I can solve the k, but I am not really able to do the rest!
The mass M of apples in grams is normally distributed with mean μ . The following table shows probabilities for values of M .
Values: M>93 -> p(x)=k
93< M < 119 -> p(x)= 0.98
119 < M -> p(x)= 0.01
Part a:
(i) Write down the value of k .
(ii) Show that μ = 106.
Part b:
Find P (M < 95) .
Part c and d:
The apples are packed in bags of ten.Any apples with a mass less than 95 g are classiied as small.
Find the probability that a bag of apples selected at random contains at most one small apple.
A crate contains 50 bags of apples. A crate is selected at random.
(i) Find the expected number of bags in this crate that contain at most one small apple.
(ii) Find the probability that at least 48 bags in this crate contain at most one small apple.
probability statistics
edited Dec 12 '18 at 13:24
David G. Stork
11k31432
asked Dec 12 '18 at 10:30
Dominik Dominik
83
$endgroup$
add a comment |
$begingroup$
I can solve the k, but I am not really able to do the rest!
The mass M of apples in grams is normally distributed with mean μ . The following table shows probabilities for values of M .
Values: M>93 -> p(x)=k
93< M < 119 -> p(x)= 0.98
119 < M -> p(x)= 0.01
Part a:
(i) Write down the value of k .
(ii) Show that μ = 106.
Part b:
Find P (M < 95) .
Part c and d:
The apples are packed in bags of ten.Any apples with a mass less than 95 g are classiied as small.
Find the probability that a bag of apples selected at random contains at most one small apple.
A crate contains 50 bags of apples. A crate is selected at random.
(i) Find the expected number of bags in this crate that contain at most one small apple.
(ii) Find the probability that at least 48 bags in this crate contain at most one small apple.
probability statistics
edited Dec 12 '18 at 13:24
David G. Stork
11k31432
asked Dec 12 '18 at 10:30
Dominik Dominik
83
$endgroup$
I can solve the k, but I am not really able to do the rest!
The mass M of apples in grams is normally distributed with mean μ . The following table shows probabilities for values of M .
Values: M>93 -> p(x)=k
93< M < 119 -> p(x)= 0.98
119 < M -> p(x)= 0.01
Part a:
(i) Write down the value of k .
(ii) Show that μ = 106.
Part b:
Find P (M < 95) .
Part c and d:
The apples are packed in bags of ten.Any apples with a mass less than 95 g are classiied as small.
Find the probability that a bag of apples selected at random contains at most one small apple.
A crate contains 50 bags of apples. A crate is selected at random.
(i) Find the expected number of bags in this crate that contain at most one small apple.
(ii) Find the probability that at least 48 bags in this crate contain at most one small apple.
probability statistics
probability statistics
edited Dec 12 '18 at 13:24
David G. Stork
11k31432
asked Dec 12 '18 at 10:30
Dominik Dominik
83
edited Dec 12 '18 at 13:24
David G. Stork
11k31432
asked Dec 12 '18 at 10:30
Dominik Dominik
83
edited Dec 12 '18 at 13:24
David G. Stork
11k31432
edited Dec 12 '18 at 13:24
David G. Stork
11k31432
edited Dec 12 '18 at 13:24
David G. Stork
11k31432
11k31432
asked Dec 12 '18 at 10:30
Dominik Dominik
83
asked Dec 12 '18 at 10:30
Dominik Dominik
83
asked Dec 12 '18 at 10:30
Dominik Dominik
83
83
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Part a) $$P(Mgt 93) = 1-P(Mlt 93) = k$$
$$P(Mlt 93) = 1-k$$
$$P(93lt M lt 119) = 0.98$$
$$P(Mgt 119) = 0.01$$
Thus $$(1-k+0.98+0.01) = 1 implies k =0.99$$
$$frac{119-mu}{sigma} = Phi^{-1}(0.99) = 2.326$$
$$frac{93-mu}{sigma} = Phi^{-1}(0.01) = -2.326$$
Adding the above we get
$$frac{119-mu}{sigma}=-frac{93-mu}{sigma}$$
$$implies mu = 106$$.
Part b)$$P(M<95) = Phi(frac{95-106}{frac{13}{2.326}}) = p$$
Part c) Atmost 1 apples be small in a bag of 10
Use Binomial Distribution to find $P(Xle1)$
$= {10choose 0} p^0 (1-p)^{10} + {10choose 1} p^1 (1-p)^9 = P$
Part d) Expected Value$ = 50*P$
$$P(Yge 48) ={50choose 48} P^{48} (1-P)^{2}+{50choose 49} P^{49}(1-P)^{1} +{50choose 50} P^{50}$$
answered Dec 12 '18 at 11:18
Satish RamanathanSatish Ramanathan
9,86531323
$endgroup$
add a comment |
$begingroup$
Due to symmetry $k=0.01$ and $mu=(119+93)/2=106$. From $P(M<119)=0.99$ you'll find $sigma=5.588$. Thus $P(M<95)=0.0245$.
For (c) and (d) use an appropriate Binomial Distribution.
edited Dec 12 '18 at 11:51
farruhota
20.2k2738
answered Dec 12 '18 at 11:19
Michael HoppeMichael Hoppe
11k31836
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Part a) $$P(Mgt 93) = 1-P(Mlt 93) = k$$
$$P(Mlt 93) = 1-k$$
$$P(93lt M lt 119) = 0.98$$
$$P(Mgt 119) = 0.01$$
Thus $$(1-k+0.98+0.01) = 1 implies k =0.99$$
$$frac{119-mu}{sigma} = Phi^{-1}(0.99) = 2.326$$
$$frac{93-mu}{sigma} = Phi^{-1}(0.01) = -2.326$$
Adding the above we get
$$frac{119-mu}{sigma}=-frac{93-mu}{sigma}$$
$$implies mu = 106$$.
Part b)$$P(M<95) = Phi(frac{95-106}{frac{13}{2.326}}) = p$$
Part c) Atmost 1 apples be small in a bag of 10
Use Binomial Distribution to find $P(Xle1)$
$= {10choose 0} p^0 (1-p)^{10} + {10choose 1} p^1 (1-p)^9 = P$
Part d) Expected Value$ = 50*P$
$$P(Yge 48) ={50choose 48} P^{48} (1-P)^{2}+{50choose 49} P^{49}(1-P)^{1} +{50choose 50} P^{50}$$
answered Dec 12 '18 at 11:18
Satish RamanathanSatish Ramanathan
9,86531323
$endgroup$
add a comment |
$begingroup$
Part a) $$P(Mgt 93) = 1-P(Mlt 93) = k$$
$$P(Mlt 93) = 1-k$$
$$P(93lt M lt 119) = 0.98$$
$$P(Mgt 119) = 0.01$$
Thus $$(1-k+0.98+0.01) = 1 implies k =0.99$$
$$frac{119-mu}{sigma} = Phi^{-1}(0.99) = 2.326$$
$$frac{93-mu}{sigma} = Phi^{-1}(0.01) = -2.326$$
Adding the above we get
$$frac{119-mu}{sigma}=-frac{93-mu}{sigma}$$
$$implies mu = 106$$.
Part b)$$P(M<95) = Phi(frac{95-106}{frac{13}{2.326}}) = p$$
Part c) Atmost 1 apples be small in a bag of 10
Use Binomial Distribution to find $P(Xle1)$
$= {10choose 0} p^0 (1-p)^{10} + {10choose 1} p^1 (1-p)^9 = P$
Part d) Expected Value$ = 50*P$
$$P(Yge 48) ={50choose 48} P^{48} (1-P)^{2}+{50choose 49} P^{49}(1-P)^{1} +{50choose 50} P^{50}$$
answered Dec 12 '18 at 11:18
Satish RamanathanSatish Ramanathan
9,86531323
$endgroup$
add a comment |
$begingroup$
Part a) $$P(Mgt 93) = 1-P(Mlt 93) = k$$
$$P(Mlt 93) = 1-k$$
$$P(93lt M lt 119) = 0.98$$
$$P(Mgt 119) = 0.01$$
Thus $$(1-k+0.98+0.01) = 1 implies k =0.99$$
$$frac{119-mu}{sigma} = Phi^{-1}(0.99) = 2.326$$
$$frac{93-mu}{sigma} = Phi^{-1}(0.01) = -2.326$$
Adding the above we get
$$frac{119-mu}{sigma}=-frac{93-mu}{sigma}$$
$$implies mu = 106$$.
Part b)$$P(M<95) = Phi(frac{95-106}{frac{13}{2.326}}) = p$$
Part c) Atmost 1 apples be small in a bag of 10
Use Binomial Distribution to find $P(Xle1)$
$= {10choose 0} p^0 (1-p)^{10} + {10choose 1} p^1 (1-p)^9 = P$
Part d) Expected Value$ = 50*P$
$$P(Yge 48) ={50choose 48} P^{48} (1-P)^{2}+{50choose 49} P^{49}(1-P)^{1} +{50choose 50} P^{50}$$
answered Dec 12 '18 at 11:18
Satish RamanathanSatish Ramanathan
9,86531323
$endgroup$
Part a) $$P(Mgt 93) = 1-P(Mlt 93) = k$$
$$P(Mlt 93) = 1-k$$
$$P(93lt M lt 119) = 0.98$$
$$P(Mgt 119) = 0.01$$
Thus $$(1-k+0.98+0.01) = 1 implies k =0.99$$
$$frac{119-mu}{sigma} = Phi^{-1}(0.99) = 2.326$$
$$frac{93-mu}{sigma} = Phi^{-1}(0.01) = -2.326$$
Adding the above we get
$$frac{119-mu}{sigma}=-frac{93-mu}{sigma}$$
$$implies mu = 106$$.
Part b)$$P(M<95) = Phi(frac{95-106}{frac{13}{2.326}}) = p$$
Part c) Atmost 1 apples be small in a bag of 10
Use Binomial Distribution to find $P(Xle1)$
$= {10choose 0} p^0 (1-p)^{10} + {10choose 1} p^1 (1-p)^9 = P$
Part d) Expected Value$ = 50*P$
$$P(Yge 48) ={50choose 48} P^{48} (1-P)^{2}+{50choose 49} P^{49}(1-P)^{1} +{50choose 50} P^{50}$$
answered Dec 12 '18 at 11:18
Satish RamanathanSatish Ramanathan
9,86531323
edited Dec 13 '18 at 0:45
answered Dec 12 '18 at 11:18
Satish RamanathanSatish Ramanathan
9,86531323
answered Dec 12 '18 at 11:18
Satish RamanathanSatish Ramanathan
9,86531323
answered Dec 12 '18 at 11:18
Satish RamanathanSatish Ramanathan
9,86531323
9,86531323
add a comment |
add a comment |
$begingroup$
Due to symmetry $k=0.01$ and $mu=(119+93)/2=106$. From $P(M<119)=0.99$ you'll find $sigma=5.588$. Thus $P(M<95)=0.0245$.
For (c) and (d) use an appropriate Binomial Distribution.
edited Dec 12 '18 at 11:51
farruhota
20.2k2738
answered Dec 12 '18 at 11:19
Michael HoppeMichael Hoppe
11k31836
$endgroup$
add a comment |
$begingroup$
Due to symmetry $k=0.01$ and $mu=(119+93)/2=106$. From $P(M<119)=0.99$ you'll find $sigma=5.588$. Thus $P(M<95)=0.0245$.
For (c) and (d) use an appropriate Binomial Distribution.
edited Dec 12 '18 at 11:51
farruhota
20.2k2738
answered Dec 12 '18 at 11:19
Michael HoppeMichael Hoppe
11k31836
$endgroup$
add a comment |
$begingroup$
Due to symmetry $k=0.01$ and $mu=(119+93)/2=106$. From $P(M<119)=0.99$ you'll find $sigma=5.588$. Thus $P(M<95)=0.0245$.
For (c) and (d) use an appropriate Binomial Distribution.
edited Dec 12 '18 at 11:51
farruhota
20.2k2738
answered Dec 12 '18 at 11:19
Michael HoppeMichael Hoppe
11k31836
$endgroup$
Due to symmetry $k=0.01$ and $mu=(119+93)/2=106$. From $P(M<119)=0.99$ you'll find $sigma=5.588$. Thus $P(M<95)=0.0245$.
For (c) and (d) use an appropriate Binomial Distribution.
edited Dec 12 '18 at 11:51
farruhota
20.2k2738
answered Dec 12 '18 at 11:19
Michael HoppeMichael Hoppe
11k31836
edited Dec 12 '18 at 11:51
farruhota
20.2k2738
edited Dec 12 '18 at 11:51
farruhota
20.2k2738
edited Dec 12 '18 at 11:51
farruhota
20.2k2738
20.2k2738
answered Dec 12 '18 at 11:19
Michael HoppeMichael Hoppe
11k31836
answered Dec 12 '18 at 11:19
Michael HoppeMichael Hoppe
11k31836
answered Dec 12 '18 at 11:19
Michael HoppeMichael Hoppe
11k31836
11k31836
add a comment |
add a comment |
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