Combine normal observations and choose a minimum sample size

Write $M\sim N(120,10^2)$. The two possible orders are disjoint, so the probability is $2P(100<M<126)P(M>130)\approx0.223$, to 3 s.f., using independent fruit selections.

$p=P(M>122)=0.4207403122\ldots$. If $X$ counts qualifying pears among 10, then $X\sim B(10,p)$ and $P(X=3)=\binom{10}{3}p^3(1-p)^7\approx0.196$, to 3 s.f.

Let independent $F_j\sim N(115,8^2)$ and $D=F_1+F_2+F_3-3M$. Then $E(D)=3(115)-3(120)=-15$ and $\operatorname{Var}(D)=3(8^2)+3^2(10^2)=1092$. Thus $D\sim N(-15,1092)$ and $P(F_1+F_2+F_3>3M)=P(D>0)\approx0.325$, to 3 s.f.

$\bar F\sim N(115,64/n)$ exactly. The condition $P(|\bar F-115|>4)\le0.3$ is equivalent to $P(-4<\bar F-115<4)\ge0.7$. Standardising gives $P(-\sqrt n/2<Z<\sqrt n/2)\ge0.7$, so $\sqrt n/2\ge\Phi^{-1}(0.85)\approx1.03643338$. Therefore $n\ge4.2968\ldots$ and the least integer is 5.

The source enters normalcdf(111,119,115,√(8²/X)) as a function of the integer sample size $X$. Its parameter screen has lower 111, upper 119, mean 115 and standard deviation $\sqrt{8^2/X}$. The printed comparison table is:

The first acceptable integer is 5. The screenshot also lists the following surrounding values; this table method is for an integer unknown.

Source screen values, to 4 decimal places: