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.
Best Answer
Part a) $$P(M\gt 93) = 1-P(M\lt 93) = k$$
$$P(M\lt 93) = 1-k$$
$$P(93\lt M \lt 119) = 0.98$$
$$P(M\gt 119) = 0.01$$ 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 $P(X\le1)$
$= {10\choose 0} p^0 (1-p)^{10} + {10\choose 1} p^1 (1-p)^9 = P$
Part d) Expected Value$ = 50*P$
$$P(Y\ge 48) ={50\choose 48} P^{48} (1-P)^{2}+{50\choose 49} P^{49}(1-P)^{1} +{50\choose 50} P^{50}$$