TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2023W/HW04.1

Miraculin

Miraculin is a protein naturally produced in a rare tropical fruit. It can convert a sour taste into a sweet taste. Consequently, miraculin has the potential to be an alternative low-calorie sweetener. A group of Japanese environmental scientists investigated the ability of a hybrid tomato plant to produce miraculin. For a particular generation of the tomato plant, the amount $X$ of miraculin produced (measured in micrograms per gram of fresh weight) had a mean 105.3 and a standard deviation of 8.0. Assume that $X$ is normally distributed. Use the table of the cdf of the standard normal distribution to compute

(a) $P(100<X<110)$

(b) $P(X(X-120)>0)$

(c) the interquartile range of the miraculin production.

Note: The interquartile range is the difference betwen the upper and

the lower quartile of $X$, i.e. $IQR_{X}=x_{0.75}-x_{0.25}$.

Dieses Beispiel ist als solved markiert. Ist dies falsch oder ungenau? Aktualisiere den Lösungsstatus (Details: Vorlage:Beispiel)

Lösungsvorschlag von Lessi

--Lessi 2024-02-07T13:04:11Z

Let $X$ be the amount of Miraculin for a generation of plants. Then $X\sim {\mathcal {N}}(105.3,8^{2})$. To standardize this random variable, we define $Z={\frac {X-105.3}{8}}\sim {\mathcal {N}}(0,1)$

a)

We want the probability o f $P(100<X<110)$:

${\begin{aligned}P(100<X<110)&=P(100\leq X\leq 110)=P(X\leq 110)-P(X\leq 100)\\&=P{\big (}{\frac {X-105.3}{8}}\leq {\frac {110-105.3}{8}}{\big )}-P{\big (}{\frac {X-105.3}{8}}\leq {\frac {100-105.3}{8}}{\big )}\\&=P{\big (}Z\leq {\frac {110-105.3}{8}}{\big )}-P{\big (}Z\leq {\frac {100-105.3}{8}}{\big )}\end{aligned}}$

z110 <- (110 - 105.3) / 8
print(z110)

z100 <- (100 - 105.3) / 8
print(z100)

So by taking a look at the table we get $P(100<X<110)=\Phi (0.5875)-\Phi (-0.6625)=0.7190-0.2546=0.4644$

# just to confirm: 
pnorm(z110) - pnorm(z100)

b)

Basically the same as before:

${\begin{aligned}P(X(X-120)>0)&=P(X^{2}>120X)=P(X>120)=1-P(X\leq 120)\\&=1-P{\big (}{\frac {X-105.3}{8}}\leq {\frac {120-105.3}{8}}{\big )}\\&=1-P{\big (}Z\leq {\frac {120-105.3}{8}}{\big )}\end{aligned}}$

z120 <- (120 - 105.3) / 8
print(z120)

Which means that $P(X>120)=1-\Phi (1.8375)=1-0.9664=0.0336$

# confirm again:
pnorm(z120, lower.tail = FALSE)

c)

First we compute the lower and upper quartile for the standard normal distribution:

${\begin{aligned}z_{0.25}&=\Phi ^{-1}(0.25)\approx -0.67\\z_{0.75}&=\Phi ^{-1}(0.75)\approx 0.67\end{aligned}}$

To get $x_{0.25}$ and $x_{0.75}$, we transform $z_{p}={\frac {x_{p}-105.3}{8}}$:

${\begin{aligned}-0.67&={\frac {x_{0.25}-105.3}{8}}\\-5.36&=x_{0.25}-105.3\\x_{0.25}&=99.94\end{aligned}}$