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vovikov84 [41]
2 years ago
9

GOD BELIEVES YOU CAN HELP ME

Mathematics
2 answers:
Paladinen [302]2 years ago
8 0

Answer:

40%

You are welcome <33

Step-by-step explanation:

12345 [234]2 years ago
3 0

Answer:

40 percent

2/5 is 40 percent

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1. Whitney Gourmet Cat Food has determined the weight of their cat food can is normally distributed with a mean of 3 ounces and
statuscvo [17]

Answer:

The probability it will weigh between 2.95 and 3.1 ounces is 0.8186.

Step-by-step explanation:

We are given that Whitney Gourmet Cat Food has determined the weight of their cat food can is normally distributed with a mean of 3 ounces and a standard deviation of 0.05 ounces.

<u><em>Let X = weight of their cat food can</em></u>

So, X ~ Normal(\mu=3,\sigma^{2} =0.05^{2})

The z score probability distribution for normal distribution is given by;

                                Z =  \frac{X-\mu}{\sigma} ~ N(0,1)

where, \mu = population mean = 3 ounces

            \sigma = standard deviation = 0.05 ounces

Now, the probability it will weigh between 2.95 and 3.1 ounces is given by = P(2.95 ounces < X < 3.1 ounces)

P(2.95 ounces < X < 3.1 ounces) = P(X < 3.1 ounces) - P(X \leq 2.95 ounces)

   P(X < 3.1 ounces) = P( \frac{X-\mu}{\sigma} < \frac{3.1-3}{0.05} ) = P(Z < 2) = 0.97725

    P(X \leq 2.95 ounces) = P( \frac{X-\mu}{\sigma} \leq \frac{2.95-3}{0.05} ) = P(Z \leq -1) = 1 - P(Z < 1)

                                                              = 1 - 0.84134 = 0.15866

The above probabilities is calculated by looking at the value of x = 2 and x = 1 in the z table which has an area of 0.97725 and 0.84134 respectively.

Therefore, P(2.95 ounces < X < 3.1 ounces) = 0.97725 - 0.15866 = 0.8186

<em>Hence, the probability it will weigh between 2.95 and 3.1 ounces is 0.8186.</em>

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