Question

In a photographic process, the developing time of prints may be looked upon a random variable having normal distribution with mean of 16.28 seconds and a standard deviation of 0.12 seconds. Find the probability that it will take:
1.anywhere from 16.00 to 16.50 seconds.
2.atleast 16.20 seconds.
3. At most 16.35 seconds, to develop one of the prints.

Answer 1

When x = 16.35 development of one of the prints is 0.7190

Given:

Normal distribution with mean is 16.28 and standard deviation of 0.12

To Find:

To get the probability that will take from 16.00 to 16.50seconds and at least 16.20 seconds

Solution:

Let x be the random variable having long the normal distribution with the mean of 16.28 seconds and a standard deviation of 0.12 seconds

Ч = 16.28, σ = 0.12

The standard normal variate is,

z = x - Ч/ σ

= $\frac{x - 16.28}{0.12}$

= 1

P(Less than 16.35 seconds) = P(X< 16.35)

So,

when x = 16.35

Z = $\frac{16.35 - 16.28}{0.12}$

= $\frac{0.17}{0.12}$

= 0.583

P(X<16.35) = P(Z<0.583)

= 0.5 + 0.2190

= 0.7190

Hence, the development of one of the prints is 0.7190

#SPJ3

Answer 2

Answer:

When $x=16.35$ the development of one of the prints is $0.7190$

Explanation:

Concept: Probability is a way to gauge how likely something is to happen. Many things are difficult to forecast with absolute confidence. Using it, we can only make predictions about how probable an event is to happen or its chance of happening. Probability can range from 0 to 1, with 0 denoting an impossibility and 1 denoting a certainty.

Given:

The normal distribution with a mean is $16.28$ and standard deviation of $0.12$

To Find:

The objective is to find out the probability that will take from $16.00$ to $16.50$seconds and at least $16.20$ seconds.

Solution:

Let x be the random variable having long the normal distribution with the mean of $16.28$ seconds and a standard deviation of $0.12$ seconds.

Ч$=16.28$, σ$=0.12$

The standard normal variate is, $z=x-$Ч/σ

$=\frac{x-16.28}{0.12}$

$=1$

P(Less than $16.35$ seconds) = P(X< $16.35$)

when $x=16.35$

$Z=\frac{16.35-16.28}{0.12} \\=0.583$

P(X<$16.35$) = P(Z<$0.583$)

$=0.5+0.2190\\=0.7190$

Therefore, the development of one of the prints is $0.7190$.

#SPJ2