Question

Oranges A home gardener likes to grow various kinds of citrus fruit. One of his

mandarin orange trees produces oranges whose circumferences follow a Normal

distribution with mean $21.1$cm and standard deviation $1.8$cm.

a. What is the probability that a randomly selected orange from this tree has a

circumference greater than $22$cm?

b. What is the probability that a random sample of $20$ oranges from this tree has a mean circumference greater than $22$ cm?

Short answer

a. The probability is $0.3085$.

b. The probability is$0.012676$

Step-by-step solution

Given Information

It is given that $\left(\mu \right)=21.1$

$\left(\sigma \right)=1.8$

$n=20$

Probability that a randomly selected orange from this tree has acircumference greater than 22 cm

Let $X$be a random variable representing circumference of orange following normal distribution having mean as $21.1$cm

Standard deviation $1.8$cm

Required probability

$P(X>22)=P\left(\frac{x-\mu }{\sigma }>\frac{22-21.1}{1.8}\right)$

$=P(Z>0.5)$

$=1-P(Z<0.5)$

$=0.3085$

Probability that a random sample of 20 oranges from this tree has a mean circumference greater than 22cm

Required probability:

$P(\overline{X}>22)=P\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}>\frac{22-21.1}{\frac{1.8}{\sqrt{20}}}\right)$

$=P(Z>2.236)$

$=1-P(Z<2.236)$

$=0.012676$