00:01
Hi everyone, welcome to the question.
00:02
So in this question, they have told us that the number of chocolate chips, an 18 ounce bag chip, is normally distributed.
00:10
And they have given us that the mean is equal to 1 to 5 -2 chips.
00:17
And the standard deviation sigma, they have given us 129 chips.
00:21
Now, they have asked us to find the probability that is a substitution probability that bag contains between, thousand and thousand four hundred chocolate chips so here we can use the formula z is equal to x minus mu divided by sigma so when x is equal to thousand it would be z is equal to thousand minus one to five two divided by sigma is one to nine so we would be getting minus one point now when x is equal to thousand four hundred we would be getting z is equal to thousand four hundred minus one two five two divided by one two nine so that is equal to one point one four seven so probability between thousand less than x less than one four zero zero is equal to p of minus one 1 .95 less than z less than 1 .47 so that is equal to p of z less than 1 .147 minus p of z less than minus 1 .95 so that is equal to from the table the value for the probability for the z score 1 .147 is 0 .87 -286 minus the probability for z score minus 1 .95 from the table is 0 .0259.
02:17
That is equal to 0 .84727.
02:21
From this, i can write the answer that probability that the randomly selected bag contains between 1 ,000 and 1 ,400 chocolate chips is equal to 0 .84727.
02:52
This is the answer for the a part.
02:55
Now, moving on to the b part, in the b part, they have asked for the probability that the bag contains fewer than thousand chocolate chips.
03:14
So that is when x is equal to thousand, the z value would be minus 1 .95.
03:21
We have seen in the a subpart.
03:23
So, p of x less than thousand is equal to p of z less than minus 1 .95.
03:31
So that is equal to 0 .0259.
03:36
So from this i can write the answer as probability that fewer than thousand chocolate chips is equal to 0 .02559 .9...