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Zepler [3.9K]
3 years ago
12

a. Suppose a BMW dealer in Fullerton, CA is trying to calculate the probability of his car sale for next week. The dealer knows

that the sale of car is normally distributed with mean 50 and variance 9. The variance 9 was calculated from the weekly car sale data of 20 weeks, as the population variance is not known to the dealer. What is the probability that the dealer will sell 51 or more cars next week? (Hint: use t distribution) (15)
Mathematics
1 answer:
egoroff_w [7]3 years ago
8 0

Answer:

0.45576

Step-by-step explanation:

z = (x-μ)/σ, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

Standard Deviation = √variance

Mean = 50

= √9

= 3

z = 51 - 50/3

= 0.11111

Probability value from Z-Table:

P(x<51) = 0.54424

P(x>51) = 1 - P(x<51)

= 1 - 0.54424

= 0.45576

The probability that the dealer will sell 51 or more cars is 0.45576

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In the supermarket a loaf of bread costs 37p how many loaves can david buy with a 2 coin ? How much money will be left over?
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6 0
3 years ago
1/2 + 5/3 - 1<br>------------------<br> 3/4
Sav [38]
Hey there :)

\frac{ \frac{1}{2}+ \frac{5}{3} -1 }{ \frac{3}{4} }
      ↓ Is the same as
\frac{1}{2} + \frac{5}{3} -1 ÷ \frac{3}{4}

Since it is a division by fraction, we can multiply by what is called the reciprocal
( The attached picture might help you )
( \frac{1}{2} + \frac{5}{3} - 1 ) × \frac{4}{3} 
     ↓ For this part, we need the common denominator, which is 6
\frac{1(3)}{2(3)}+ \frac{5(2)}{32)}  \frac{1(6)}{1(6)} × \frac{4}{3}

\frac{3}{6}+ \frac{10}{6}  - \frac{6}{6} × \frac{4}{3}
\frac{7}{6} × \frac{4}{3}
\frac{14}{9} =  1\frac{5}{9}
  ↑
Your final answer
 

7 0
3 years ago
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