Question

A particular type of mouse's weights are normally distributed, with a mean of 359 grams and a standard deviation of 33 grams. If you pick one mouse at random, find the following: (round all probabilities to four decimal places)
a) What is the probability that the mouse weighs less than 405 grams?
b) What is the probability that the mouse weighs more than 461 grams?
c) What is the probability that the mouse weighs between 406 and 461 grams?
d) Is it unlikely that a randomly chosen mouse would weigh less than 405 grams?
i. No, the likelihood exceeds 50%
ii. No, the likelihood exceeds 5%
iii. Yes, the likelihood is less than 50%
iv. Yes, the likelihood is less than 5%
e) What is the cutoff for the heaviest 10% of this type of mouse?

Answer 1

Answer:

Following are the responses to the given points:

Step-by-step explanation:

Given:

$\mu = 359 \ \ \ \ \ \sigma = 33$

Using formula:

$\to P(X = x) = P\left ( z < \frac{x-\mu }{\sigma } \right )$

For point a:

$\to x= 406\\\\P(X < 406 ) = P\left ( z < \frac{406-359 }{33 } \right )$

                   $= P\left ( z <1.4242\right )\\\\= 0.9222$

For point b:

$\to x= 461\\\\P(X > 461 ) = P\left ( z > \frac{461 -359 }{33 } \right )$

                   $= 1 - P\left ( z < 3.0909 \right )\\\\= 1 - 0.9990\\\\= 0.0001$

For point c:

$\to x= 406\ \ and \ \ 461\\\\P(406 <X < 461 ) = P(X < 461) - P (X < 406)$

                              $= 0.9990 - 0.9222\\\\= 0.0768$

For point d:

$P(X < 406) = 0.9222 = 92.22 \%\ that \ is \ greater \ than\ 50\%$

Hence, the correct choice is "i".

For point e:

$\to P(X > z) = 0.1\\\\\to 1 - P(X < z) = 0.1\\\\\to P(\frac{x-\mu }{\sigma }) = 0.9\\\\\to \frac{x-359}{33 } = 1.28\\\\ \to x = 401.24$